A
Theory & Concepts
Principles, Steps, Strong Induction, Common Mistakes
1. What is Mathematical Induction?
Definition
Mathematical Induction is a method of proof used to establish that a statement \(P(n)\) is true for all natural numbers \(n\) (or for all \(n \geq n_0\)). It is analogous to a domino effect: if the first domino falls and every falling domino knocks the next one, then all dominoes will fall.
2. Principle of Mathematical Induction (PMI)
Formal Statement of PMI
Let \(P(n)\) be a statement involving a natural number \(n\). If:
1. Base Step: \(P(1)\) (or \(P(n_0)\)) is true, AND
2. Inductive Step: Whenever \(P(k)\) is true for some \(k \in \mathbb{N}\), then \(P(k+1)\) is also true
Then \(P(n)\) is true for all \(n \in \mathbb{N}\). The assumption "\(P(k)\) is true" is called the Inductive Hypothesis (I.H.).
3. Steps to Apply PMI — The 4-Step Method
STEP 1
Write the Statement
Let \(P(n)\) be the statement to prove.
STEP 2
Base Case — Verify \(P(1)\)
Substitute \(n=1\); verify LHS = RHS. Write: "P(1) is true."
STEP 3
Inductive Hypothesis — Assume \(P(k)\)
Write: "Assume \(P(k)\) is true, i.e., [statement with \(n\) replaced by \(k\)]."
STEP 4
Inductive Step — Prove \(P(k+1)\)
Using I.H., prove \(P(k+1)\) is true. Then conclude by PMI.
4. Strong (Second) Mathematical Induction
Strong Induction
Base Step: Verify \(P(n_0)\) (sometimes multiple base cases).
Inductive Step: Assume \(P(m)\) true for all \(m\) with \(n_0 \leq m \leq k\), then prove \(P(k+1)\).
When to use: When \(P(k+1)\) requires earlier cases too. Example: Fibonacci: \(a_n = a_{n-1} + a_{n-2}\).
5. Common Mistakes to Avoid
⚠ Common Errors — Read Carefully
- Forgetting the base case — Always verify \(P(1)\) first.
- Circular reasoning — Do NOT assume \(P(k+1)\) to prove \(P(k)\).
- Wrong inductive step — Must show \(P(k) \Rightarrow P(k+1)\).
- Wrong base case — If statement starts at \(n=2\), verify \(P(2)\).
- Not using I.H. — I.H. must be explicitly used in Step 4.
B
Complete Formula Sheet
Summation · AP & GP · Identities · Partial Fractions
6. Standard Summation Formulas (Must Memorise)
Basic Summation Formulas
Formula (1)
\(\displaystyle\sum_{r=1}^n r = \frac{n(n+1)}{2}\)Formula (2)
\(\displaystyle\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}\)Formula (3)
\(\displaystyle\sum_{r=1}^n r^3 = \left[\frac{n(n+1)}{2}\right]^2\)Formula (4)
\(\displaystyle\sum_{r=1}^n r^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\)Formula (5)
\(1+3+5+\cdots+(2n-1) = n^2\)Formula (6)
\(2+4+6+\cdots+2n = n(n+1)\)Product & Combined Summation Formulas
Formula (7)
\(\displaystyle\sum_{r=1}^n r(r+1) = \frac{n(n+1)(n+2)}{3}\)Formula (8)
\(\displaystyle\sum_{r=1}^n r(r+1)(r+2) = \frac{n(n+1)(n+2)(n+3)}{4}\)Formula (9)
\(\displaystyle\sum_{r=1}^n r(r+2) = \frac{n(n+1)(2n+7)}{6}\)Formula (10)
\(\displaystyle\sum_{r=1}^n (2r-1)^2 = \frac{n(2n-1)(2n+1)}{3}\)Formula (11)
\(1^2+3^2+\cdots+(2n-1)^2 = \frac{n(2n-1)(2n+1)}{3}\)Formula (12)
\(1^3+3^3+\cdots+(2n-1)^3 = n^2(2n^2-1)\)7. AP & GP Formulas
Arithmetic Progression (AP)
\(T_n = a + (n-1)d\)
\(S_n = \dfrac{n}{2}[2a+(n-1)d] = \dfrac{n}{2}(a+l)\)
Key: \(a,b,c\) in AP \(\Rightarrow 2b = a+c\)
Geometric Progression (GP)
\(T_n = ar^{n-1}\)
\(S_n = \dfrac{a(r^n-1)}{r-1}\), \quad S_\infty = \dfrac{a}{1-r}\) \((|r|<1)\)
Key: \(a,b,c\) in GP \(\Rightarrow b^2 = ac\)
8. Algebraic Identities Used in Induction
Binomial & Algebraic Identities
\((a+b)^2 = a^2+2ab+b^2\)
\((a-b)^2 = a^2-2ab+b^2\)
\((a+b)^3 = a^3+3a^2b+3ab^2+b^3\)
\((a-b)^3 = a^3-3a^2b+3ab^2-b^3\)
\(a^n - b^n\) always divisible by \((a-b)\)
\(a^n + b^n\) divisible by \((a+b)\) when \(n\) is odd
9. Partial Fractions — Telescoping Method
Telescoping Sum Formula
\(\dfrac{1}{(a+(r-1)d)\cdot(a+rd)} = \dfrac{1}{d}\left(\dfrac{1}{a+(r-1)d} - \dfrac{1}{a+rd}\right)\)
After telescoping: \(\displaystyle S_n = \frac{1}{d}\left(\frac{1}{a_1} - \frac{1}{a_{n+1}}\right)\)
Common Telescoping Results (EAPCET Favourites)
Series
\(\frac{1}{1\cdot5}+\cdots\) to \(n\) terms \(= \dfrac{n}{4n+1}\)Series
\(\frac{1}{2\cdot5}+\cdots\) to \(n\) terms \(= \dfrac{n}{6n+4}\)Series
\(\frac{1}{3\cdot7}+\cdots\) to \(n\) terms \(= \dfrac{n}{3(4n+3)}\)Series
\(\frac{1}{1\cdot7}+\cdots\) to \(n\) terms \(= \dfrac{n}{6(6n+1)}\)C
Topic-wise Solved Examples
5 Types · 13 Fully Worked Examples from Past EAPCET
Ex 1
AP/TG EAPCET 2015
Prove: \(1\times2\times3 + 2\times3\times4 + \cdots\) to \(n\) terms \(= \dfrac{n(n+1)(n+2)(n+3)}{4}\)
Step 1 (n=1):LHS \(= 6\); RHS \(= \frac{1\cdot2\cdot3\cdot4}{4} = 6\) ✓
Step 2 (I.H.):\(\displaystyle\sum_{r=1}^k r(r+1)(r+2) = \frac{k(k+1)(k+2)(k+3)}{4}\)
Step 3 (Need):\(\displaystyle\sum_{r=1}^{k+1} r(r+1)(r+2) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}\)
Step 4:LHS \(= \frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) = (k+1)(k+2)(k+3)\left(\frac{k}{4}+1\right) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}\) ✓
✓ By PMI, P(n) is true for all n ∈ ℕ
Ex 3
AP EAPCET 2024
Find: \(2 + 3 + 5 + 6 + 8 + 9 + \cdots + 2n\) terms
Strategy:Group in pairs: \((2+3),(5+6),(8+9),\ldots\) giving sums \(5,11,17,\ldots\) (AP with \(a=5, d=6\), \(n\) pairs)
Sum:\(S = \frac{n}{2}(6n+4) = n(3n+2) = 3n^2+2n\)
Verify (n=1):\(2+3=5\); formula \(= 3+2=5\) ✓
Answer: (a) \(3n^2+2n\)
Ex 4
TS EAPCET 2021
Prove \(9 \mid (4^n + 15n - 1)\) for all \(n \in \mathbb{N}\).
Step 1 (n=1):\(4+15-1 = 18 = 9\times2\) ✓
I.H.:Assume \(4^k+15k-1 = 9m\), so \(4^k = 9m-15k+1\)
Step 4:\(4^{k+1}+15(k+1)-1 = 4\cdot4^k+15k+14 = 4(9m-15k+1)+15k+14 = 36m-60k+4+15k+14 = 9(4m-5k+2)\) ✓
✓ 9 | (4ⁿ + 15n − 1) for all n ∈ ℕ by PMI
Ex 5
TS EAPCET 2024
Largest divisor of \(81^n + 20^n - 1\) is \(k\). Sum of divisors is \(S\). Find \(S-k\).
n=1:\(100\); n=2: \(6600\); gcd\((100, 6600) = 100 = 2^2 \times 5^2\). So \(k = 100\).
Divisors:\(1,2,4,5,10,20,25,50,100\); \quad S = 217
Answer: (a) S − k = 117
Ex 7
TS EAPCET 2025
Greatest integer \(k\) such that \(2^{n+4}+12 \geq k(n+4)\) for all \(n \in \mathbb{N}\).
Substitution:Let \(m = n+4 \geq 5\). Need \(k \leq \frac{2^m+12}{m}\) for all \(m\geq5\).
Check m=5:\(\frac{44}{5} = 8.8\); m=6: 12.67; m=7: 20. Tightest bound at m=5 gives \(k \leq 8.8\).
Verify n=1:\(44 \geq 40\) ✓
Answer: (b) k = 8
Ex 9
TS EAPCET 2021
\(\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+\cdots\) to \(n\) terms \(= \dfrac{27}{109}\). Find \(n\).
Formula:AP: 1,5,9,…; \(d=4\). \(T_r = \frac{1}{4}\left(\frac{1}{4r-3}-\frac{1}{4r+1}\right)\). \(S_n = \frac{n}{4n+1}\)
Solve:\(\frac{n}{4n+1} = \frac{27}{109}\) \(\Rightarrow\) \(109n = 108n+27\) \(\Rightarrow\) \(n=27\)
Answer: (b) n = 27
Ex 11
AP EAPCET 2024
\(1\cdot3\cdot5 + 3\cdot5\cdot7 + \cdots\) to \(n\) terms \(= n(n+1)f(n) - 3n\). Find \(f(1)\).
n=1:LHS \(= 1\cdot3\cdot5 = 15\); RHS \(= 2f(1)-3 = 15\) \(\Rightarrow\) \(f(1) = 9\)
Answer: (a) f(1) = 9
Ex 13
AP EAPCET 2024
\(2\cdot5 + 5\cdot9 + 8\cdot13 + 11\cdot17 + \cdots\) to 10 terms.
General:\(a_r = 3r-1\); \(b_r = 4r+1\); \(T_r = 12r^2-r-1\)
Sum:\(S_{10} = 12\cdot\frac{10\cdot11\cdot21}{6} - \frac{10\cdot11}{2} - 10 = 4620-55-10 = 4555\)
Answer: (b) 4555
D
Quick Reference Card
Important Results · EAPCET Pattern · High-Yield Tips
Summation Formulas
\(\sum r = \frac{n(n+1)}{2}\)
\(\sum r^2 = \frac{n(n+1)(2n+1)}{6}\)
\(\sum r^3 = \left[\frac{n(n+1)}{2}\right]^2\)
\(\sum r(r+1) = \frac{n(n+1)(n+2)}{3}\)
\(\sum r(r+1)(r+2) = \frac{n(n+1)(n+2)(n+3)}{4}\)
Divisibility Key Results
\(4^n+15n-1 \equiv 0 \pmod{9}\)
\(3^{2n}-1 \equiv 0 \pmod{8}\)
\(3^{2n+2}-8n-9 \equiv 0 \pmod{64}\)
\(2^{2n+1}+3^{2n+1} \equiv 0 \pmod{5}\)
\(3(5^{2n+1})+2^{3n+1} \equiv 0 \pmod{17}\)
\(49^n+16^n-1 \equiv 0 \pmod{64}\)
\(81^n+20^n-1 \equiv 0 \pmod{100}\)
⭐ EAPCET High-Yield Tips
Greatest divisor: Compute gcd(f(1), f(2))
Telescoping: Memorise formula; verify n=1
f(n) problems: Just substitute the given n
Factors of P \(= p_1^{e_1}\cdots\): Count \((e_1+1)(e_2+1)\cdots\)
Sum of divisors: \(\sigma(p^a) = \frac{p^{a+1}-1}{p-1}\)
Most asked: Telescoping (9+/year), Divisibility, f(n) type
36 Topic-wise Practice Questions
AP/TG EAPCET 2021–2025 · Previous Year Questions
TOPIC 1
Sum of Series by Induction
Q1–Q10
Q1
2 May 2025
Shift-I
Shift-I
\(1+(1+3)+(1+3+5)+(1+3+5+7)+\cdots\) to 10 terms =
Q2
AP May 22, 2024 (I)
\(2+3+5+6+8+9+\cdots+2n\) terms =
Q3
10 Sep 2020, Shift-II
If \(S_n\) is sum of first \(n\) terms of \(1^2+2\cdot2^2+3^2+2\cdot4^2+5^2+\cdots\), then when \(n\) is even, \(S_n =\)
Q4
4 May 2019
\(n+2(n-1)+3(n-2)+\cdots+(n-1)\cdot2+n\cdot1 = \alpha n(n+1)(n+2)\), then \(\alpha =\)
Q5
AP Apr 21, 2019 (I)
\(1^2+(1^2+2^2)+(1^2+2^2+3^2)+\cdots+(1^2+\cdots+n^2) =\)
Q6
AP 2015
\(1\times2\times3+2\times3\times4+3\times4\times5+\cdots\) up to \(n\) terms =
Q7
TS 2017
For any integer \(n \geq 1\), \(\displaystyle\sum_{k=1}^n k(k+2) =\)
Q8
3 May 2025, Shift-II
Greatest integer \(k\) satisfying \(2^{n+4}+12 \geq k(n+4)\) for all \(n \in \mathbb{N}\) is
Q9
TS May 10, 2024 (I)
Among the following, which is NOT true for all \(n \in \mathbb{N}\)?
Q10
3 May 2019 (II)
For all \(n \in \mathbb{N}\), if \(1^2+2^2+\cdots+n^2 > x\), then \(x =\)
TOPIC 2
Divisibility Results by Induction
Q11–Q23
Q112 May 2025, Shift-I
If \(5^{2n}-48n+k\) is divisible by 24 for all \(n \in \mathbb{N}\), least positive \(k\) is
Q124 May 2025, Shift-II
If \(2^{4n+3}+3^{3n+1}\) is divisible by \(P\) for all \(n \in \mathbb{N}\), then \(P\) is
Q13TS Aug 4, 2021 (II)
For all \(n \in \mathbb{N}\), \(2^{2n+1}+3^{2n+1}\) is divisible by
Q14TS Aug 6, 2021 (I)
For any \(n \in \mathbb{N}\), \(4^n+15n-1\) is divisible by
Q15TS Aug 6, 2021 (II)
For all \(n \in \mathbb{N}\), \(3(5^{2n+1})+2^{3n+1}\) is divisible by
Q16TS May 10, 2024 (II)
\(3^{2n+2}-8n-9\) is divisible by \(2^p\), \(\forall n \in \mathbb{N}\). Maximum value of \(p\) is
Q17TS Sep 11, 2020 (I)
\(n^5-5n^3+4n\) is divisible by 120 for
Q184 May 2019
Greatest divisor of \(30\cdot5^{2n}+4\cdot2^{3n}\) is \(p\) and of \(2^{2n+1}-6n-2\) is \(q\). Then \(p+q =\)
Q196 May 2019
If \(\alpha, \beta\) are least positive integers such that \(3|n^3+\alpha n\) and \(6|n^3-\beta n\) for all \(n\), then \(\alpha+\beta =\)
Q20AP EAPCET 2024
If \(P\) is greatest divisor of \(49^n+16^n-1\) for all \(n \in \mathbb{N}\), number of factors of \(P\) is
Q21TS May 9, 2024 (I)
Largest divisor of \(81^n+20^n-1\) is \(k\). If \(S\) = sum of all positive divisors of \(k\), then \(S-k =\)
Q22AP EAPCET 2024
If \(2\cdot4^{2n+1}+3^{3n+1}\) is divisible by \(k\) for all \(n \in \mathbb{N}\), then \(k =\)
Q23AP May 18, 2024 (I)
For all positive integers \(n\), if \(3(5^{2n+1})+2^{3n+1}\) is divisible by \(k\), number of primes \(\leq k\) is
TOPIC 3
Telescoping, Partial Fractions & Product Series
Q24–Q36
Q243 May 2025, Shift-I
\(\frac{1}{2\cdot7}+\frac{1}{7\cdot12}+\frac{1}{12\cdot17}+\frac{1}{17\cdot22}+\cdots\) to 10 terms \(= k\). Then \(k =\)
Q25TS Sep 10, 2020 (I)
gcd\((m,n)=1\). If \(\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\cdots\) (20 terms) \(=\frac{m}{n}\), then \(5m+2n =\)
Q26TS May 9, 2024 (II)
\(\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+\frac{1}{9\cdot12}+\cdots\) to 9 terms \(=\)
Q27TS Aug 5, 2021 (II)
If \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\cdots\) to \(n\) terms \(=\frac{27}{109}\), then \(n =\)
Q28TS Aug 4, 2021 (I)
For all \(n \in \mathbb{N}\), \(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\cdots+\frac{1}{(3n-1)(3n+2)} =\)
Q29AP May 21, 2024 (I)
\(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\cdots\) up to \(n\) terms =
Q30AP May 20, 2024 (I)
\(\frac{1}{3\cdot7}+\frac{1}{7\cdot11}+\frac{1}{11\cdot15}+\cdots\) to 50 terms =
Q31AP 2017
Sum of first \(n\) terms of \(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots =\)
Q32AP 2019 (S)
For the series \(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\cdots = \frac{n}{6n+4}\), sum to 12 terms is
Q33AP May 19, 2024 (II)
\(2\cdot5+5\cdot9+8\cdot13+11\cdot17+\cdots\) to 10 terms =
Q34AP May 21, 2024 (II)
If \(1\cdot3\cdot5+3\cdot5\cdot7+\cdots\) to \(n\) terms \(= n(n+1)f(n)-3n\), then \(f(1) =\)
Q35TS May 11, 2024 (I)
If \(1\cdot3\cdot5+3\cdot5\cdot7+\cdots\) to \(n\) terms \(= n(n+1)f(n)\), then \(f(2) =\)
Q36AP May 23, 2024 (I)
The \(n\)th term of \(1+(3+5+7)+(9+11+13+15+17)+\cdots\) is
Answer Key — All 36 Questions
Q1
(a)
Q2
(a)
Q3
(c)
Q4
(d)
Q5
(c)
Q6
(d)
Q7
(b)
Q8
(b)
Q9
(c)
Q10
(a)
Q11
(d)
Q12
(a)
Q13
(b)
Q14
(b)
Q15
(b)
Q16
(c)
Q17
(a)
Q18
(b)
Q19
(b)
Q20
(c)
Q21
(a)
Q22
(b)
Q23
(c)
Q24
(c)
Q25
(c)
Q26
(c)
Q27
(b)
Q28
(a)
Q29
(c)
Q30
(b)
Q31
(c)
Q32
(b)
Q33
(b)
Q34
(a)
Q35
(b)
Q36
(d)
S
Detailed Solutions — All 36 Questions
Step-by-step workings for every question
Q1
\(r\)-th group \(= 1+3+\cdots+(2r-1) = r^2\). So \(S = \displaystyle\sum_{r=1}^{10}r^2 = \frac{10\cdot11\cdot21}{6} = 385\)
Answer: (a) 385
Q2
Pairs: \((2+3),(5+6),(8+9),\ldots\) — AP \(a=5, d=6\), \(n\) pairs. \(S = \frac{n}{2}(6n+4) = 3n^2+2n\)
Answer: (a) \(3n^2+2n\)
Q3
\(k\)-th pair: \((2k-1)^2+2(2k)^2 = 12k^2-4k+1\). With \(m=n/2\), summing gives \(S_n = \frac{n(n+1)^2}{2}\). Verify \(n=2\): \(1+2\cdot4=9 = \frac{2\cdot9}{2}\) ✓
Answer: (c) \(\frac{n(n+1)^2}{2}\)
Q4
\(T_r = r(n-r+1)\). \(S = (n+1)\frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(n+2)}{6}\). So \(\alpha = \frac{1}{6}\)
Answer: (d) \(\frac{1}{6}\)
Q5
\(T_r = \frac{r(r+1)(2r+1)}{6}\). \(\displaystyle S = \frac{1}{6}\sum(2r^3+3r^2+r) = \frac{n(n+1)^2(n+2)}{12}\). Verify \(n=2\): \(\frac{2\cdot9\cdot4}{12}=6\) ✓
Answer: (c) \(\frac{n(n+1)^2(n+2)}{12}\)
Q6
\(\displaystyle\sum_{r=1}^n r(r+1)(r+2) = \frac{n(n+1)(n+2)(n+3)}{4}\). Verify \(n=1\): \(6=6\) ✓
Answer: (d) \(\frac{n(n+1)(n+2)(n+3)}{4}\)
Q7
\(\displaystyle\sum(k^2+2k) = \frac{n(n+1)(2n+1)}{6}+n(n+1) = \frac{n(n+1)(2n+7)}{6}\)
Answer: (b) \(\frac{n(n+1)(2n+7)}{6}\)
Q8
\(m=n+4\geq5\); need \(k\leq\frac{2^m+12}{m}\). At \(m=5\): \(\frac{44}{5}=8.8\) (tightest). So \(k=8\).
Answer: (b) 8
Q9
Check (c) for \(n=2\): \(3\cdot5^5+2^7 = 9375+128=9503\). \(9503\div23 = 413.17...\) — NOT divisible by 23. (It's divisible by 17.)
Answer: (c) — Not true
Q10
\(\frac{n(n+1)(2n+1)}{6} > \frac{n^3}{3}\) iff \((n+1)(2n+1) > 2n^2\) iff \(3n+1>0\) — always true.
Answer: (a) \(\frac{n^3}{3}\)
Q11
\(n=1\): \(25-48+k = k-23\). Need \(24|(k-23)\). Least \(k=23\). Verify \(n=2\): \(552=24\times23\) ✓
Answer: (d) 23
Q12
\(n=1\): \(128+81=209=11\times19\). \(n=2\): \(2048+2187=4235=11^2\times35\). gcd\(=11\) (odd prime). So \(P=11\).
Answer: (a) even integer (Note: The answer key says (a); 11 is odd prime, answer key may differ from option labeling)
Q13
\(n=1\): \(8+27=35=5\times7\). \(n=2\): \(32+243=275=5^2\times11\). gcd\(=5\). Also \(2^{2n+1}+3^{2n+1}\) is \(a^n+b^n\) with odd power, divisible by \(2+3=5\).
Answer: (b) 5
Q14
\(n=1\): \(18=9\times2\). \(n=2\): \(45=9\times5\). gcd\(=9\). Also \(4^n=(1+3)^n\Rightarrow 4^n+15n-1 = 9(\ldots)\).
Answer: (b) 9
Q15
\(n=1\): \(375+16=391=17\times23\). \(n=2\): \(9375+128=9503=17\times559\). gcd\(=17\).
Answer: (b) 17
Q16
\(n=1\): \(81-17=64=2^6\). \(n=2\): \(729-25=704=2^6\times11\). Max \(p=6\).
Answer: (c) 6
Q17
\(n^5-5n^3+4n = n(n^2-1)(n^2-4) = (n-2)(n-1)n(n+1)(n+2)\) — product of 5 consecutive integers, always divisible by \(5!=120\).
Answer: (a) all positive integers
Q18
\(p\): \(n=1\): \(782\); \(n=2\): \(19006\); gcd\(=34\). \(q\): \(n=2\): \(18\); \(n=3\): \(108\); gcd\(=18\). \(p+q=52\).
Answer: (b) 52
Q19
\(\alpha\): \(n=1\): \(1+\alpha\equiv0\pmod3\Rightarrow\alpha=2\). \(\beta\): \(n=1\): \(1-\beta\equiv0\pmod6\Rightarrow\beta=1\). \(\alpha+\beta=3\).
Answer: (b) 3
Q20
\(n=1\): \(64=2^6\). \(n=2\): \(2432=2^7\times19\). gcd\(=64=2^6\). \(P=64\). Factors: \(6+1=7\).
Answer: (c) 7
Q21
\(n=1\): \(100\); \(n=2\): \(6600\); gcd\(=100=2^2\times5^2\). \(k=100\). \(S=1+2+4+5+10+20+25+50+100=217\). \(S-k=117\).
Answer: (a) 117
Q22
\(n=1\): \(128+81=209=11\times19\). \(n=2\): \(2048+2187=4235=11^2\times35\). gcd\(=11\). \(k=11\).
Answer: (b) 11
Q23
From Q15: divisor \(k=17\). Primes \(\leq17\): 2,3,5,7,11,13,17 — total 7 primes.
Answer: (c) 7
Q24
AP: 2,7,12,…; \(d=5\). \(S_{10}=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{52}\right) = \frac{1}{5}\cdot\frac{50}{104} = \frac{5}{52}\)
Answer: (c) \(\frac{5}{52}\)
Q25
AP: 1,7,13,…; \(d=6\). \(S_n = \frac{n}{6n+1}\). \(S_{20}=\frac{20}{121}\). gcd\((20,121)=1\). \(5(20)+2(121)=342\).
Answer: (c) 342
Q26
\(\frac{1}{3r\cdot3(r+1)} = \frac{1}{9r(r+1)}\). \(S_9 = \frac{1}{9}\left(1-\frac{1}{10}\right) = \frac{1}{10}\)
Answer: (c) \(\frac{1}{10}\)
Q27
\(d=4\). \(S_n = \frac{n}{4n+1} = \frac{27}{109}\). \(109n=108n+27\Rightarrow n=27\).
Answer: (b) 27
Q28
\(d=3\). \(S_n = \frac{1}{3}\left(\frac{1}{2}-\frac{1}{3n+2}\right) = \frac{n}{2(3n+2)} = \frac{n}{6n+4}\)
Answer: (a) \(\frac{n}{6n+4}\)
Q29
\(d=4\). \(S_n = \frac{1}{4}\left(\frac{1}{1}-\frac{1}{4n+1}\right) = \frac{n}{4n+1}\)
Answer: (c) \(\frac{n}{4n+1}\)
Q30
\(d=4\). \(S_n = \frac{n}{3(4n+3)}\). \(S_{50} = \frac{50}{3\times203} = \frac{50}{609}\)
Answer: (b) \(\frac{50}{609}\)
Q31
\(d=3\). \(S_n = \frac{1}{3}\left(\frac{1}{2}-\frac{1}{3n+2}\right) = \frac{n}{2(3n+2)}\)
Answer: (c) \(\frac{n}{2(3n+2)}\)
Q32
\(S_{12} = \frac{12}{6(12)+4} = \frac{12}{76} = \frac{3}{19}\)
Answer: (b) \(\frac{3}{19}\)
Q33
\(a_r=3r-1\), \(b_r=4r+1\). \(T_r=12r^2-r-1\). \(S_{10}=12(385)-55-10=4620-65=4555\)
Answer: (b) 4555
Q34
\(n=1\): LHS \(=1\cdot3\cdot5=15\); RHS \(=2f(1)-3=15\Rightarrow f(1)=9\)
Answer: (a) 9
Q35
\(n=2\): LHS \(=15+3\cdot5\cdot7=15+105=120=6f(2)\Rightarrow f(2)=20\)
Answer: (b) 20
Q36
Group \(n\) has \(2n-1\) terms; first term of group \(n\) is \(2(n-1)^2+1\). Sum of group \(n = (2n-1)[(n-1)^2+n^2]\). Check: \(T_1=1\), \(T_2=15\), \(T_3=65\) ✓
Answer: (d) \((2n-1)[(n-1)^2+n^2]\)