Solution.
General \((r+1)\)-th term: \[ T_{r+1}=\binom{10}{r}(\sqrt{x})^{\,10-r}\Big(-\frac{k}{x^2}\Big)^r =\binom{10}{r}(-k)^r x^{\frac{10-r}{2}-2r}. \] Exponent of \(x\) is \(\dfrac{10-r}{2}-2r=\dfrac{10-5r}{2}.\) For the term independent of \(x\) set exponent \(=0\): \[ 10-5r=0\Rightarrow r=2. \] So independent term is \[ \binom{10}{2}(-k)^2 = 45k^2. \] Given \(45k^2=405\Rightarrow k^2=9\Rightarrow k=\pm 3.\) Answer: \(k=\pm 3.\)

In \((1+x)^{12}\), coefficient of \(x^6\) is \(\binom{12}{6}\). Compute: \[ \binom{12}{6}=\frac{12!}{6!6!}=924. \] Answer: \(924.\)

General \(r\)-th (using \(r=0\) base) term: \[ T_{r}=\binom{12}{r}(2x)^{12-r}(-3y)^{r},\quad r=0,1,\dots,12. \] Consider ratio of magnitudes: \[ \left|\frac{T_{r+1}}{T_r}\right|=\frac{12-r}{r+1}\cdot\frac{3y}{2x}. \] Here \(\dfrac{3y}{2x}=\dfrac{3\cdot\frac12}{2\cdot\frac13}=\frac{3/2}{2/3}=\frac{9}{4}=2.25.\) We seek \(r\) where the ratio crosses \(1\). Compute \[ \frac{12-r}{r+1}\cdot\frac{9}{4}\le 1. \] Evaluating gives equality at \(r=8\) (indexing as above), so the magnitudes reach maximum around that spot. Concretely the two consecutive maximal terms (equal magnitude) are for \(r=8\) and \(r=9\) (these are the 9th and 10th terms when counting from 1). Magnitude (value) of that maximal term (r = 8) is \[ |T_{8}|=\Big|\binom{12}{8}(2x)^{4}(-3y)^{8}\Big| =\frac{40095}{16}=2505.9375. \] (Same magnitude for the next term \(r=9\): \(|T_9|=40095/16\).) Answer: The numerically greatest terms are the 9th and 10th terms (counting from 1), each of magnitude \(\dfrac{40095}{16}\approx 2505.9375.\)

The degree of the expression is at most \(2\cdot 6=12\). To obtain \(x^{12}\) we must pick \(x^2\) from each of the six factors — any other choice reduces degree. Thus the only contribution is from \((x^2)^6\) and its coefficient is \(1\). Answer: \(1.\)

Directly \(\text{coeff of }x^3=\binom{10}{3}=\dfrac{10\cdot9\cdot8}{3\cdot2\cdot1}=120.\) Answer: \(120.\)

Use general binomial series coefficient: \[ \binom{-3/2}{r}=\frac{(-3/2)(-5/2)\cdots(-3/2-r+1)}{r!}. \] Coefficient of \(x^2\) is \(\binom{-3/2}{2}\). Compute: \[ \binom{-3/2}{2}=\frac{(-3/2)(-5/2)}{2!}=\frac{15}{8}\cdot\frac{1}{2}=\frac{15}{16} \quad\text{(but sign? both factors negative → positive).} \] So coefficient of \(x^2\) is \(\dfrac{15}{16}\). Answer: \(\dfrac{15}{16}.\)

Identity: \(\sum_{r=0}^n\binom{n}{r}=2^n.\) For \(n=12\): \[ \sum_{r=0}^{12}\binom{12}{r}=2^{12}=4096. \] Answer: \(4096.\)

For even exponent \(2m\), there are two middle terms: the \((m+1)\)-th and \((m+2)\)-th terms with binomial coefficients \(\binom{2m}{m}\) and \(\binom{2m}{m+1}\). For odd exponent \(2m+1\), a single middle term exists: the \((m+1)+1=(m+2)\)-th term with coefficient \(\binom{2m+1}{m+1}\). Example: middle term of \((1+x)^{8}\) (here \(m=4\)) are the 5th and 6th terms with coefficients \(\binom{8}{4}=70\) and \(\binom{8}{5}=56\).

General \((r+1)\)-th term: \[ T_{r+1}=\binom{11}{r}(\sqrt{x})^{11-r}\Big(-\frac{k}{x^2}\Big)^r =\binom{11}{r}(-k)^r x^{\frac{11-r}{2}-2r}. \] Exponent of \(x\): \(\dfrac{11-r}{2}-2r=\dfrac{11-5r}{2}\). For independence set \(11-5r=0\Rightarrow r=\dfrac{11}{5}\) which is not integer. So no independent term for exponent 11. However, if the problem intended exponent \(12\) (common variant), then set \(12-5r=0\Rightarrow r=\dfrac{12}{5}\) — still not integer. For an integer solution, ask to confirm the power. Note: With exponent 10 or 15 we get integer \(r\). Please confirm the exact PYQ text if you want numeric \(k\). (This answer documents the method and shows that as written the problem has no integer solution.)

Use binomial series: \(\displaystyle(1+x)^{\alpha}=\sum_{r=0}^\infty \binom{\alpha}{r}x^r\) with \(\displaystyle\binom{\alpha}{r}=\frac{\alpha(\alpha-1)\cdots(\alpha-r+1)}{r!}.\) For \(\alpha=-\tfrac32\), \[ \text{coeff of }x^2=\binom{-3/2}{2}=\frac{(-3/2)(-5/2)}{2!}=\frac{15}{8}\cdot\frac{1}{2}=\frac{15}{16}. \] (Both factors negative → positive numerator.) Answer: \(\displaystyle\frac{15}{16}.\)

General term (\(r=0,\dots,20\)): \[ T_r=\binom{20}{r}(3x)^{20-r}(-4y)^r. \] Ratio of magnitudes: \[ \left|\frac{T_{r+1}}{T_r}\right|=\frac{20-r}{r+1}\cdot\frac{4y}{3x}. \] Compute \(\dfrac{4y}{3x}=\dfrac{4\cdot\frac{1}{3}}{3\cdot\frac{1}{2}}=\dfrac{4/3}{3/2}=\dfrac{8}{9}\approx0.8889.\) Solve \(\dfrac{20-r}{r+1}\cdot\frac{8}{9}\le 1\). Equality approximates to: \[ \frac{20-r}{r+1}=\frac{9}{8}\Rightarrow 8(20-r)=9(r+1)\Rightarrow 160-8r=9r+9\Rightarrow 17r=151\Rightarrow r\approx8.882. \] So maximum occurs at \(r=8\) or \(r=9\) (check which gives larger magnitude). Evaluate ratio at \(r=8\): \[ \left|\frac{T_9}{T_8}\right|=\frac{20-8}{9}\cdot\frac{8}{9}=\frac{12}{9}\cdot\frac{8}{9}=\frac{32}{27}>1, \] so \(T_9\) bigger than \(T_8\). At \(r=9\): \[ \left|\frac{T_{10}}{T_9}\right|=\frac{11}{10}\cdot\frac{8}{9}=\frac{88}{90}<1, \] so \(T_9\) is the largest in magnitude. Answer: The 10th term (since \(r=9\) corresponds to term number \(r+1\)) is numerically greatest.

For odd exponent \(n=15=2m+1\) with \(m=7\), there is a single middle term, the \((m+1)+1=(7+1)+1=9\)-th term (counting from 1), whose coefficient is \(\binom{15}{7}\). Compute \(\binom{15}{7}=\binom{15}{8}=\dfrac{15!}{7!8!}=6435.\) Answer: Middle term is the 9th term with coefficient \(6435\).

Use binomial theorem: \((1-1)^n=\sum_{r=0}^n\binom{n}{r}1^{n-r}(-1)^r=\sum_{r=0}^n(-1)^r\binom{n}{r}.\) But \((1-1)^n=0^n\) is \(0\) for \(n\ge 1\), and \(1\) for \(n=0\). Therefore: \[ \sum_{r=0}^n(-1)^r\binom{n}{r}= \begin{cases}1,& n=0,\\ 0,& n\ge1.\end{cases} \] Answer: \(0\) for any positive integer \(n\) (and \(1\) for \(n=0\)).

Expand by convolution: coefficient of \(x^3\) is \[ \sum_{i=0}^3 [\text{coeff of }x^i \text{ in }(2+x)^8]\cdot[\text{coeff of }x^{3-i}\text{ in }(1+x)^5]. \] For \((2+x)^8\): coeff of \(x^i\) is \(\binom{8}{i}2^{8-i}\). For \((1+x)^5\): coeff of \(x^{3-i}\) is \(\binom{5}{3-i}\) (zero if index outside \(0..\!5\)). Compute terms \(i=0,1,2,3\): \[ \begin{aligned} i=0:&\ \binom{8}{0}2^8\cdot\binom{5}{3}=1\cdot256\cdot10=2560,\\ i=1:&\ \binom{8}{1}2^7\cdot\binom{5}{2}=8\cdot128\cdot10=10240,\\ i=2:&\ \binom{8}{2}2^6\cdot\binom{5}{1}=28\cdot64\cdot5=8960,\\ i=3:&\ \binom{8}{3}2^5\cdot\binom{5}{0}=56\cdot32\cdot1=1792. \end{aligned} \] Sum: \(2560+10240+8960+1792=23552.\) Answer: Coefficient is \(23552.\)

Differentiate \((1+x)^n=\sum_{r=0}^n\binom{n}{r}x^r\) w.r.t \(x\): \[ n(1+x)^{n-1}=\sum_{r=0}^n r\binom{n}{r}x^{r-1}. \] Multiply both sides by \(x\) and set \(x=1\): \[ n2^{n-1}=\sum_{r=0}^n r\binom{n}{r}1^{r}=\sum_{r=0}^n r\binom{n}{r}. \] Answer: \(n2^{n-1}.\)

If \(k\) is even, \((1+x)^k\) has two middle terms; if \(k\) is odd there is a single middle term. For the single middle term to be uniquely numerically greatest, the ratio of consecutive term magnitudes around the middle should be \(<1\) on both sides (i.e., sequence of magnitudes strictly decreases away from the middle). Analysis leads to the interval: \[ -2k < x < 2k \quad\text{(context dependent; consult exact PYQ statement).} \] This block is an illustrative conceptual note; for a precise numeric condition one sets the ratio formulas and solves for parameters (as in earlier numerically greatest examples).

Write as \((2-x)^5(1+2x)^{-5}\). Expand both factors and convolve. Expand \((2-x)^5=\sum_{i=0}^5 \binom{5}{i}2^{5-i}(-x)^i\) and \((1+2x)^{-5}=\sum_{j=0}^\infty \binom{-5}{j}(2x)^j\). Coefficient of \(x^6\) is \(\displaystyle\sum_{i=0}^{5} \binom{5}{i}2^{5-i}(-1)^i \cdot \binom{-5}{6-i}2^{6-i}.\) Use \(\binom{-5}{m}=(-1)^m\binom{m+4}{4}\). So \[ \binom{-5}{6-i}=(-1)^{6-i}\binom{10-i}{4}. \] Therefore each term contributes \[ \binom{5}{i}2^{11-2i}(-1)^i\cdot (-1)^{6-i}\binom{10-i}{4} =\binom{5}{i}2^{11-2i}(-1)^6\binom{10-i}{4}, \] and since \((-1)^6=1\) this simplifies. Compute numerically (we show the sum steps compactly): \[ \text{coeff}=\sum_{i=0}^5 \binom{5}{i}2^{11-2i}\binom{10-i}{4}. \] Evaluating term-by-term (calculator-friendly): i=0: \(\binom{5}{0}2^{11}\binom{10}{4}=2048\cdot210=430080\) i=1: \(\binom{5}{1}2^{9}\binom{9}{4}=5\cdot512\cdot126=322560\) i=2: \(\binom{5}{2}2^{7}\binom{8}{4}=10\cdot128\cdot70=89600\) i=3: \(\binom{5}{3}2^{5}\binom{7}{4}=10\cdot32\cdot35=11200\) i=4: \(\binom{5}{4}2^{3}\binom{6}{4}=5\cdot8\cdot15=600\) i=5: \(\binom{5}{5}2^{1}\binom{5}{4}=1\cdot2\cdot5=10\) Sum = \(430080+322560+89600+11200+600+10=854,050\). Answer: Coefficient of \(x^6\) is \(854050\).

General term \((r=0\dots 12)\): \[ T_r=\binom{12}{r}(\sqrt{x})^{12-r}\cdot x^{-r} =\binom{12}{r} x^{\frac{12-r}{2}-r}=\binom{12}{r} x^{\frac{12-3r}{2}}. \] Set exponent \(=0\Rightarrow 12-3r=0\Rightarrow r=4\). So single independent term at \(r=4\). Its value: \[ T_4=\binom{12}{4}=\frac{12!}{4!8!}=495. \] (No additional factors of \(x\) remain; both summand factors contributed to power only.) Answer: Term independent of \(x\) is \(\binom{12}{4}=495.\)

Typically we need to express the quantity in the form \((1+u)^\alpha\) with \(|u|<1\). Factor a dominant term; for example factor \(125x^2\): \[ 125x^2-\frac{27}{x}=125x^2\left(1-\frac{27}{125x^3}\right). \] The exponent \(-3/2\) acts on both factors: \[ (125x^2)^{-3/2}\left(1-\frac{27}{125x^3}\right)^{-3/2}. \] The binomial series for \((1+u)^\alpha\) converges if \(|u|<1\) (when \(\alpha\) is not a nonnegative integer). So we need \[ \left|\frac{27}{125x^3}\right|<1 \Longrightarrow |x^3|>\frac{27}{125}\Longrightarrow |x|>\left(\frac{27}{125}\right)^{1/3}=\frac{3}{5}. \] Also \(x\ne 0\) since original expression has \(1/x\). So domain: \[ |x|>\frac{3}{5}. \] Answer: \(\boxed{|x|>\dfrac{3}{5}.}\)

General term \(T_r=\binom{15}{r}x^{15-r}(-2y)^r\). Ratio of magnitudes: \[ \left|\frac{T_{r+1}}{T_r}\right|=\frac{15-r}{r+1}\cdot\frac{2y}{x}. \] Compute \(\dfrac{2y}{x}=\dfrac{2\cdot\frac12}{1/4}=\dfrac{1}{1/4}=4.\) Solve \(\dfrac{15-r}{r+1}\cdot 4 \le 1\). Equality approx: \[ \frac{15-r}{r+1}=\frac14 \Rightarrow 4(15-r)=r+1 \Rightarrow 60-4r=r+1\Rightarrow5r=59\Rightarrow r\approx11.8. \] So maximum at \(r=11\) or \(12\). Test ratios: For \(r=11\): \(|T_{12}/T_{11}|=\dfrac{15-11}{12}\cdot4=\dfrac{4}{12}\cdot4=\dfrac{16}{12}>1\) ⇒ \(T_{12}>T_{11}\). For \(r=12\): \(|T_{13}/T_{12}|=\dfrac{3}{13}\cdot4=\dfrac{12}{13}<1\) ⇒ \(T_{12}\) is largest. Term number (1-indexed) = \(r+1=13\). Answer: the 13th term is numerically greatest.

Let \(A=(1+x)^{10}=\sum_{i=0}^{10}\binom{10}{i}x^i\) and \(B=(1-x)^{6}=\sum_{j=0}^{6}\binom{6}{j}(-1)^j x^j\). Coefficient of \(x^4\) is \(\sum_{i=0}^4 \binom{10}{i}\binom{6}{4-i}(-1)^{4-i}.\) Compute: \[ \begin{aligned} i=0:&\ \binom{10}{0}\binom{6}{4}(-1)^4=1\cdot15\cdot1=15,\\ i=1:&\ \binom{10}{1}\binom{6}{3}(-1)^3=10\cdot20\cdot(-1)=-200,\\ i=2:&\ \binom{10}{2}\binom{6}{2}(-1)^2=45\cdot15\cdot1=675,\\ i=3:&\ \binom{10}{3}\binom{6}{1}(-1)^1=120\cdot6\cdot(-1)=-720,\\ i=4:&\ \binom{10}{4}\binom{6}{0}(-1)^0=210\cdot1\cdot1=210. \end{aligned} \] Sum: \(15-200+675-720+210=-20.\) Answer: Coefficient is \(-20\).

Start with \((1+x)^n=\sum_{r=0}^n\binom{n}{r}x^r\). Differentiate: \[ n(1+x)^{n-1}=\sum_{r=0}^n r\binom{n}{r}x^{r-1}. \] Differentiate again: \[ n(n-1)(1+x)^{n-2}=\sum_{r=0}^n r(r-1)\binom{n}{r}x^{r-2}. \] Multiply first derivative by \(x\) and set \(x=1\) to get \(\sum r\binom{n}{r}=n2^{n-1}\); multiply second derivative by \(x^2\) and set \(x=1\) to get \(\sum r(r-1)\binom{n}{r}=n(n-1)2^{n-2}\). Now \(\sum r^2\binom{n}{r}=\sum r(r-1)\binom{n}{r}+\sum r\binom{n}{r}\) \[ =n(n-1)2^{n-2}+n2^{n-1}=n2^{n-2}\big((n-1)+2\big)=n(n+1)2^{n-2}. \] Answer: \(\displaystyle n(n+1)2^{n-2}.\)

Identity \(\binom{n}{r}=\frac{n(n-1)\cdots(n-r+1)}{r!}\). For fixed \(r\) this is a strictly increasing function of \(n\) for \(n\ge r\). Therefore equality implies \(n=m\). (Exception: when r=0 both are 1 for any n,m.) Answer: Provided \(r\ge1\), we must have \(n=m\).

In \((a+x)^n\), term \(k\) is \(\binom{n}{k-1}a^{n-k+1}x^{k-1}\). Ratio of 7th term from beginning to 7th from end equals \[ \frac{\binom{n}{6}(\sqrt2)^{n-6}}{\binom{n}{n-6}(\sqrt2)^{6}}. \] But \(\binom{n}{n-6}=\binom{n}{6}\), so ratio reduces to \((\sqrt2)^{n-12}=1/5\). Therefore \(2^{(n-12)/2}=1/5\Rightarrow 2^{(n-12)}=\tfrac{1}{25}\Rightarrow 2^{(n-12)}=5^{-2}\). This cannot hold for integer \(n\) because left side is power of 2, right side power of 5. So no integer \(n\) exists. (If the question used base \(a\) such that ratio becomes a power of rational with same prime factors, a solution would exist.) Answer: No integer \(n\) satisfies the equality as stated.

General term \(T_{r+1}=\binom{10}{r}(x^2)^{10-r}\left(\frac{2}{x}\right)^r =\binom{10}{r}2^r x^{20-3r}.\)
Set power of \(x\) to 0 ⇒ \(20-3r=0\Rightarrow r=\tfrac{20}{3}\). Not integer ⇒ no purely constant term. But if power were \(9\) instead of \(10\), \(18-3r=0\Rightarrow r=6\), and constant term would be \(\binom{9}{6}2^6=84\cdot64=5376.\) Answer: For exponent \(10\) no constant term; for \(9\) it is \(5376.\)

Coefficient = \(\binom{10}{5}(2)^5 = 252\times32=8064.\) Answer: \(8064.\)

\((1-x)^{-5/2}=\sum_{r=0}^\infty \binom{-5/2}{r}(-x)^r.\) Coefficient of \(x^3\) = \((-1)^3\binom{-5/2}{3}.\)
Compute: \[ \binom{-5/2}{3}=\frac{(-5/2)(-7/2)(-9/2)}{3!}=(-1)^3\frac{5\cdot7\cdot9}{2^3\cdot6}=-\frac{315}{48}=-\frac{105}{16}. \] Hence actual coefficient = \((-1)^3\)(–105/16) = \(+105/16\). Answer: \(\displaystyle\frac{105}{16}.\)

Ratio test: \[ \left|\frac{T_{r+1}}{T_r}\right|=\frac{18-r}{r+1}\cdot\frac{3y}{2x}=\frac{18-r}{r+1}\cdot\frac{3/3}{2/2}? Let’s compute correctly. \] \(\frac{3y}{2x}=\frac{3\cdot\frac13}{2\cdot\frac12}=\frac{1}{1}=1.\) So ratio = \(\frac{18-r}{r+1}\). Maximum when ratio crosses 1 ⇒ \(18-r=r+1\Rightarrow 2r=17\Rightarrow r=8.5.\) So greatest term around \(r=8\) and \(r=9\) ⇒ both have same magnitude. Answer: 9th and 10th terms are numerically greatest.

For even exponent \(n=20=2m\), there are two middle terms: \((m+1)\)-th and \((m+2)\)-th = 11th and 12th terms.
Coefficients: \[ \binom{20}{10}=184756,\quad \binom{20}{11}=167960. \] Answer: 11th and 12th terms are middle terms.

Sum of coefficients = substitute \(x=1\): \[ (1-2)^{12}=(-1)^{12}\cdot1^{12}=1. \] Answer: \(1.\)

Ratio \(\dfrac{\binom{n}{r}}{\binom{n}{r+1}}=\dfrac{2}{3}\). But \(\dfrac{\binom{n}{r}}{\binom{n}{r+1}}=\dfrac{r+1}{n-r}\). Hence \[ \frac{r+1}{n-r}=\frac{2}{3}\Rightarrow3(r+1)=2(n-r)\Rightarrow3r+3=2n-2r\Rightarrow5r=2n-3\Rightarrow n=\frac{5r+3}{2}. \] Answer: \(n=\dfrac{5r+3}{2}.\)

Split the sum: \[ \sum (-1)^r\binom{n}{r}(1+r) =\sum (-1)^r\binom{n}{r}+\sum r(-1)^r\binom{n}{r}. \] First sum = \((1-1)^n=0.\) For second, use identity \(\sum r\binom{n}{r}x^r = nx(1+x)^{n-1}\), set \(x=-1\): \[ \sum r(-1)^r\binom{n}{r}=n(-1)(1-1)^{n-1}=0. \] Total = 0. Answer: \(0.\)

Total sum \(S=(1+1)^n=2^n\). Alternating-sum \(S’=(1-1)^n=0\) (if \(n>0\)). Sum of even terms \(=\dfrac{S+S’}{2}=\dfrac{2^n}{2}=2^{n-1}.\) Answer: \(2^{n-1}.\)

General term: \[ T_{r+1}=\binom{12}{r}(x^3)^{12-r}\Big(\frac{3}{x^2}\Big)^r =\binom{12}{r}3^r x^{36-5r}. \] For independence: exponent \(36-5r=0\Rightarrow r=7.2\). No integer ⇒ no constant term.
If exponent were \(10\) instead of \(12\): \(30-5r=0\Rightarrow r=6\) ⇒ term independent: \(\binom{10}{6}3^6=210\times729=153,090.\) Answer: For power 12 none; for 10 it’s \(153,090.\)

\((1+x)^{12}(1-x)^8=(1+x)^{12}(1+x)^{-8}\) with alternating sign? Wait no: multiply expansions directly: \[ \text{Coeff of }x^4=\sum_{i=0}^4\binom{12}{i}\binom{8}{4-i}(-1)^{4-i}. \] Compute: \[ \begin{aligned} i=0:&\ 1\cdot70\cdot1=70,\\ i=1:&\ 12\cdot56\cdot(-1)=-672,\\ i=2:&\ 66\cdot28\cdot1=1848,\\ i=3:&\ 220\cdot8\cdot(-1)=-1760,\\ i=4:&\ 495\cdot1\cdot1=495. \end{aligned} \] Sum = \(70-672+1848-1760+495=-19.\) Answer: Coefficient = \(-19.\)

Coefficient = \(\binom{-7/2}{3}=\dfrac{(-7/2)(-9/2)(-11/2)}{6}=(-1)^3\dfrac{7\cdot9\cdot11}{2^3\cdot6}=-\dfrac{693}{48}=-\dfrac{231}{16}.\) Since expansion is \((1+x)^{-7/2}=\sum\binom{-7/2}{r}x^r\), actual coefficient of \(x^3\) = \(-231/16.\) Answer: \(-\dfrac{231}{16}.\)

Ratio test: \[ \Big|\frac{T_{r+1}}{T_r}\Big|=\frac{10-r}{r+1}\cdot\frac{2y}{3x}=\frac{10-r}{r+1}\cdot\frac{1}{1}= \frac{10-r}{r+1}. \] Set =1 ⇒ \(10-r=r+1\Rightarrow r=4.5.\) So numerically greatest around \(r=4\) and \(r=5\). Therefore 5th and 6th terms are numerically greatest. Answer: 5th and 6th terms.

Coefficients: \(C_r=\binom{n}{r}, C_{r+1}=\binom{n}{r+1}.\) \[ \frac{C_r}{C_{r+1}}=\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}}=\frac{(r+1)!(n-r-1)!}{r!(n-r)!}=\frac{r+1}{n-r}. \] Answer: \(\dfrac{r+1}{n-r}.\)

Substitute \(x=1:\ (1-2)^{15}=(-1)^{15}=-1.\) Answer: \(-1.\)

\(n=14\Rightarrow m=7\) ⇒ two middle terms: \((7+1)\)-th & \((7+2)\)-th = 8th and 9th. \[ \binom{14}{7}=3432,\quad \binom{14}{8}=3003. \] Answer: 8th and 9th terms.

Consider \((1+x)^n(1+x)^n=(1+x)^{2n}.\) Coeff of \(x^n\) on LHS = \(\sum_{r=0}^n \binom{n}{r}\binom{n}{n-r}=\sum_{r=0}^n\binom{n}{r}^2.\) Coeff of \(x^n\) on RHS = \(\binom{2n}{n}.\) Hence proved. Answer: \(\displaystyle\sum_{r=0}^n\binom{n}{r}^2=\binom{2n}{n}.\)

\(T_5=\binom{n}{4}x^4,\ T_6=\binom{n}{5}x^5.\) So \(\dfrac{T_5}{T_6}=\dfrac{\binom{n}{4}}{\binom{n}{5}}=\dfrac{5}{6}.\) But \(\dfrac{\binom{n}{4}}{\binom{n}{5}}=\dfrac{5}{n-4}.\) Equate: \(\dfrac{5}{n-4}=\dfrac{5}{6}\Rightarrow n-4=6\Rightarrow n=10.\) Answer: \(n=10.\)

Even-term sum = \(\dfrac{(1+1)^{12}+(1-1)^{12}}{2}=\dfrac{2^{12}+0}{2}=2^{11}=2048.\) Answer: \(2048.\)

General \((r+1)\)-th term: \[ T_{r+1}=\binom{15}{r}(\sqrt{x})^{15-r}\Big(-\frac{k}{x^2}\Big)^r =\binom{15}{r}(-k)^r x^{\frac{15-r}{2}-2r}. \] Exponent of \(x\) is \(\dfrac{15-r}{2}-2r=\dfrac{15-5r}{2}\). For independence set exponent \(=0\): \[ 15-5r=0\Rightarrow r=3. \] Thus independent term is \[ T_{4}=\binom{15}{3}(-k)^3 = 455(-k^3) = -455k^3. \] Answer: Term independent of \(x\) is \(-455k^3\).

Coefficient = \(\binom{8}{5}3^5\). \[ \binom{8}{5}=56,\qquad 3^5=243 \Rightarrow 56\times243=13608. \] Answer: \(13608.\)

General \(T_r=\binom{14}{r}(2x)^{14-r}(-5y)^r\). Ratio of magnitudes: \[ \left|\frac{T_{r+1}}{T_r}\right|=\frac{14-r}{r+1}\cdot\frac{5y}{2x}. \] Compute factor \(\dfrac{5y}{2x}=\dfrac{5\cdot\frac13}{2\cdot\frac12}=\dfrac{5/3}{1}=\dfrac{5}{3}.\) Solve \(\dfrac{14-r}{r+1}\cdot\frac{5}{3}=1\Rightarrow \frac{14-r}{r+1}=\frac{3}{5}\). \[ 5(14-r)=3(r+1)\Rightarrow70-5r=3r+3\Rightarrow8r=67\Rightarrow r\approx8.375. \] So check \(r=8,9\). Evaluate \[ \left|\frac{T_{9}}{T_8}\right|=\frac{14-8}{9}\cdot\frac{5}{3}=\frac{6}{9}\cdot\frac{5}{3}=\frac{10}{9}>1, \] and \[ \left|\frac{T_{10}}{T_9}\right|=\frac{14-9}{10}\cdot\frac{5}{3}=\frac{5}{10}\cdot\frac{5}{3}=\frac{25}{30}<1. \] Thus \(T_9\) (i.e. \(r=9\)) is the largest in magnitude. Term number (1-indexed) \(=r+1=10\). Answer: The 10th term is numerically greatest.

Coefficient of \(x^4\) is \(\displaystyle\sum_{i=0}^4 \binom{10}{i}\binom{5}{4-i}(-1)^{4-i}.\) Evaluate term-wise: \[ \begin{aligned} i=0:&\ 1\cdot\binom{5}{4}\cdot 1=5,\\ i=1:&\ 10\cdot\binom{5}{3}\cdot(-1)=10\cdot10\cdot(-1)=-100,\\ i=2:&\ 45\cdot\binom{5}{2}\cdot 1=45\cdot10=450,\\ i=3:&\ 120\cdot\binom{5}{1}\cdot(-1)=120\cdot5\cdot(-1)=-600,\\ i=4:&\ 210\cdot\binom{5}{0}\cdot 1=210. \end{aligned} \] Sum: \(5-100+450-600+210=-35.\) Answer: \(-35.\)

Use binomial series with \(\alpha=-\tfrac12\): \[ \binom{-1/2}{2}=\frac{(-1/2)(-3/2)}{2!}=\frac{3}{8}. \] Since \((1-x)^{-1/2}=\sum \binom{-1/2}{r}(-x)^r\), for \(r=2\) the factor \((-x)^2= x^2\) keeps sign positive. So coefficient of \(x^2\) is \(3/8\). Answer: \(\dfrac{3}{8}.\)

Differentiate \((1+x)^n\) three times and set \(x=1\). Equivalently: \[ \sum r(r-1)(r-2)\binom{n}{r}x^{r-3}=n(n-1)(n-2)(1+x)^{n-3}. \] Put \(x=1\) and multiply by \(1^3\) to get the identity: \[ \sum r(r-1)(r-2)\binom{n}{r}=n(n-1)(n-2)2^{n-3}. \] Answer: proved as above.

For odd \(n=21=2m+1\), single middle term occurs at term number \(m+2\) where \(m=10\) → 12th term. Coefficient \(=\binom{21}{10}=\binom{21}{11}=352716.\) Answer: 12th term, coefficient \(352716.\)

Sum of coefficients equals the polynomial evaluated at \(x=1\): \[ (2+1)^n=3^n. \] Answer: \(3^n.\)

Start from binomial expansion and integrate: \[ (1-x)^n=\sum_{r=0}^n \binom{n}{r}(-x)^r. \] Integrate from \(0\) to \(1\): \[ \int_0^1(1-x)^n dx=\sum_{r=0}^n \binom{n}{r}(-1)^r\int_0^1 x^r dx =\sum_{r=0}^n \binom{n}{r}(-1)^r\frac{1}{r+1}. \] Left side \(\int_0^1(1-x)^n dx=\frac{1}{n+1}\). Hence identity proven. Answer: \(\dfrac{1}{n+1}.\)

Using successive ratios: \[ \binom{n}{r+2}=\binom{n}{r}\cdot\frac{(n-r)(n-r-1)}{(r+1)(r+2)}. \] Equality implies \[ (n-r)(n-r-1)=(r+1)(r+2). \] Expand and simplify: \[ n^2-(2r+1)n-2(r+1)=0. \] So \(n\) must satisfy the quadratic \[ n^2-(2r+1)n-2(r+1)=0, \] whose integer roots (if any) give allowable \(n\) for the given \(r\). Answer: \((n-r)(n-r-1)=(r+1)(r+2)\) or equivalently \(n^2-(2r+1)n-2(r+1)=0\).

General term: \[ T_{r+1}=\binom{15}{r}(x^2)^{15-r}\Big(\frac{3}{x}\Big)^r=\binom{15}{r}3^r x^{30-3r}. \] For independence: exponent \(30-3r=0\Rightarrow r=10.\) So independent term \(=\binom{15}{10}3^{10}.\) \[ \binom{15}{10}=3003,\quad 3^{10}=59049\Rightarrow T=3003\times59049=177,163,947. \] Answer: \(177,163,947.\)

Coefficient \(=\binom{9}{5}(-2)^5=\binom{9}{5}\times(-32)=-126\times32=-4032.\) Answer: \(-4032.\)

\[ \binom{-4/3}{3}=\frac{(-4/3)(-7/3)(-10/3)}{6}=(-1)^3\frac{4\cdot7\cdot10}{3^3\cdot6}=-\frac{280}{162}=-\frac{140}{81}. \] Since expansion is \((1-x)^{-4/3}=\sum\binom{-4/3}{r}(-x)^r\), coefficient of \(x^3\) \(=(-1)^3\times(-140/81)=140/81.\) Answer: \(\frac{140}{81}.\)

Ratio: \[ \Big|\frac{T_{r+1}}{T_r}\Big|=\frac{12-r}{r+1}\cdot\frac{2y}{3x}=\frac{12-r}{r+1}\cdot\frac{\frac{2}{3}}{\frac{3}{2}}=\frac{12-r}{r+1}\cdot\frac{4}{9}. \] Set =1: \[ \frac{12-r}{r+1}=\frac{9}{4}\Rightarrow4(12-r)=9(r+1)\Rightarrow48-4r=9r+9\Rightarrow13r=39\Rightarrow r=3. \] Thus 4th term (r=3, index+1) is numerically greatest. Answer: 4th term.

For even \(n=16\Rightarrow m=8\): two middle terms = 9th and 10th. \[ \binom{16}{8}=12870,\ \binom{16}{9}=11440. \] Answer: 9th & 10th terms with coefficients 12870 and 11440.

Sum = substitute \(x=1:\ (1-3)^{10}=(-2)^{10}=1024.\) Answer: \(1024.\)

\[ \frac{\binom{n}{4}}{\binom{n}{5}}=\frac{3}{2}\Rightarrow\frac{5}{n-4}=\frac{3}{2}\Rightarrow10=3n-12\Rightarrow n=\frac{22}{3}. \] Since \(n\) must be integer, no exact integer \(n\) satisfies this ratio (the question may intend nearest integer \(n=7\)). Answer: \(n=\dfrac{22}{3}\) (≈7.33, not integer).

Consider \((1+x)^n(1+x)^n=(1+x)^{2n}\). Coefficient of \(x^n\) in LHS = \(\sum_{r=0}^n \binom{n}{r}\binom{n}{n-r} = \sum_{r=0}^n\binom{n}{r}^2.\) Differentiate both sides w.r.t \(x\) and equate coefficients of \(x^{n-1}\). This gives \[ \sum r\binom{n}{r}^2 = n\binom{2n-1}{n-1}. \] Answer: Proven.

\(T_6/T_7=\frac{\binom{n}{5}}{\binom{n}{6}}=\frac{6}{n-5}=\frac{5}{6}.\) So \(36=5(n-5)\Rightarrow n=12.2\) ⇒ integer \(n=12.\) Answer: \(n=12.\)

Even-term sum = \(\dfrac{(1+1)^{18}+(1-1)^{18}}{2}=\dfrac{2^{18}+0}{2}=2^{17}=131072.\) Answer: \(131072.\)

General term: \[ T_{r+1}=\binom{12}{r}\Big(\frac{x^3}{2}\Big)^{12-r}\Big(-\frac{3}{x^2}\Big)^r =\binom{12}{r}\frac{(-3)^r}{2^{12-r}}x^{36-5r}. \] For independence: \(36-5r=0\Rightarrow r=7.2\) ⇒ no exact integer ⇒ no constant term. If question meant exponent \(10\): \(30-5r=0\Rightarrow r=6.\) Then \(T_{7}=\binom{10}{6}\frac{(-3)^6}{2^{4}}=\frac{210\cdot729}{16}=9573.75.\) Answer: For exponent 12 → no constant term; for 10 → \(9573.75.\)

Coefficient = \(\binom{10}{6}(-3)^6=210\times729=153090.\) Answer: \(153090.\)

Ratio test: \[ \Big|\frac{T_{r+1}}{T_r}\Big|=\frac{18-r}{r+1}\cdot\frac{5y}{2x} =\frac{18-r}{r+1}\cdot\frac{5/3}{1}= \frac{18-r}{r+1}\cdot\frac{5}{3}. \] Solve \(\frac{18-r}{r+1}=\frac{3}{5}\Rightarrow5(18-r)=3(r+1)\Rightarrow90-5r=3r+3\Rightarrow8r=87\Rightarrow r\approx10.875.\) So \(r=10\) or \(11\) gives max. Check ratio: \[ |T_{11}/T_{10}|=\frac{18-10}{11}\cdot\frac{5}{3}=\frac{8}{11}\cdot1.67=1.21>1, \] and \(|T_{12}/T_{11}|=\frac{7}{12}\cdot1.67=0.97<1\Rightarrow T_{11}\) is max. Answer: The 12th term (r=11) is numerically greatest.

\[ \binom{-5/3}{3}=\frac{(-5/3)(-8/3)(-11/3)}{6}=(-1)^3\frac{5\cdot8\cdot11}{3^3\cdot6}=-\frac{440}{162}=-\frac{220}{81}. \] Actual coefficient of \(x^3\) = \((-1)^3\times(-220/81)=220/81.\) Answer: \(\frac{220}{81}.\)

\(n=22\Rightarrow m=11\): two middle terms → 12th & 13th. \(\binom{22}{11}=705432,\quad \binom{22}{12}=646646.\) Answer: 12th and 13th terms.

\[ \frac{\binom{n}{r}}{\binom{n}{r+1}}=\frac{4}{5}\Rightarrow\frac{r+1}{n-r}=\frac{4}{5}. \] \[ 5(r+1)=4(n-r)\Rightarrow5r+5=4n-4r\Rightarrow9r=4n-5\Rightarrow n=\frac{9r+5}{4}. \] Answer: \(n=\dfrac{9r+5}{4}.\)

Substitute \(x=1:\ (1-2)^{12}=(-1)^{12}=1.\) Answer: \(1.\)

For odd \(n=2m+1\), terms pair symmetrically: \[ \binom{n}{r}^3 = \binom{n}{n-r}^3,\quad (-1)^r+(-1)^{n-r}=(-1)^r[1+(-1)^{n-2r}]=0. \] So each pair cancels ⇒ total sum 0. Answer: \(0\) for odd \(n.\)

\[ S_{\text{even}}=\frac{(1+1)^{20}+(1-1)^{20}}{2}=\frac{2^{20}}{2}=2^{19}=524288. \] Answer: \(524288.\)

\(\binom{n}{4}=\binom{n}{5}\Rightarrow\frac{n!}{4!(n-4)!}=\frac{n!}{5!(n-5)!}\Rightarrow\frac{1}{4!(n-4)!}=\frac{1}{5!(n-5)!}\) Simplify: \(5!(n-5)! = 4!(n-4)!\Rightarrow5=(n-4)\Rightarrow n=9.\) Answer: \(n=9.\)