Aimstutorial

Let \(P(n) = 2^{4n+3} + 3^{3n+1}\). We will test for \(n=1\) and \(n=2\) to find a common divisor.

Step 1: Base Case (n=1)
\(P(1) = 2^{4(1)+3} + 3^{3(1)+1} = 2^7 + 3^4\)
\(P(1) = 128 + 81 = 209\)

Step 2: Case (n=2)
\(P(2) = 2^{4(2)+3} + 3^{3(2)+1} = 2^{11} + 3^7\)
\(P(2) = 2048 + 2187 = 4235\)

Step 3: Find Common Divisors
Let’s find the factors of 209.
\(209 = 11 \times 19\).
Now let’s check if 4235 is divisible by 11 or 19.
\(4235 \div 11 = 385\). (Divisible)
\(4235 \div 19 = 222.89…\) (Not divisible)

The common divisor is 11. Let’s prove \(P(n)\) is divisible by 11 for all \(n \in N\) by induction.

Step 4: Inductive Hypothesis
Assume \(P(k) = 2^{4k+3} + 3^{3k+1}\) is divisible by 11.
So, \(2^{4k+3} + 3^{3k+1} = 11m\) for some integer \(m\).
\(\implies 2^{4k+3} = 11m – 3^{3k+1}\)

Step 5: Inductive Step (Prove for n=k+1)
We need to prove that \(P(k+1) = 2^{4(k+1)+3} + 3^{3(k+1)+1}\) is divisible by 11.
\(P(k+1) = 2^{4k+4+3} + 3^{3k+3+1}\)
\(P(k+1) = 2^{4k+7} + 3^{3k+4}\)
\(P(k+1) = 2^4 \cdot 2^{4k+3} + 3^3 \cdot 3^{3k+1}\)
\(P(k+1) = 16 \cdot (2^{4k+3}) + 27 \cdot (3^{3k+1})\)

Substitute \(2^{4k+3}\) from Step 4:
\(P(k+1) = 16 \cdot (11m – 3^{3k+1}) + 27 \cdot (3^{3k+1})\)
\(P(k+1) = 16(11m) – 16 \cdot 3^{3k+1} + 27 \cdot 3^{3k+1}\)
\(P(k+1) = 16(11m) + 3^{3k+1} \cdot (27 – 16)\)
\(P(k+1) = 16(11m) + 3^{3k+1} \cdot (11)\)
\(P(k+1) = 11 \cdot (16m + 3^{3k+1})\)

Since \(P(k+1)\) is a multiple of 11, the statement is true for \(n=k+1\).
Therefore, the expression is divisible by 11 for all \(n \in N\).

The number 11 is an odd prime integer.

Let \(P(n) = 3^{2n+2} – 8n – 9\).

Step 1: Base Case (n=1)
\(P(1) = 3^{2(1)+2} – 8(1) – 9\)
\(P(1) = 3^4 – 8 – 9 = 81 – 17 = 64\).
Since 64 is divisible by 64, the statement is true for \(n=1\).

Step 2: Base Case (n=2)
\(P(2) = 3^{2(2)+2} – 8(2) – 9\)
\(P(2) = 3^6 – 16 – 9 = 729 – 25 = 704\).
\(704 \div 64 = 11\). So, \(P(2)\) is also divisible by 64.

Step 3: Inductive Hypothesis
Assume \(P(k) = 3^{2k+2} – 8k – 9\) is divisible by 64.
So, \(3^{2k+2} – 8k – 9 = 64m\) for some integer \(m\).
\(\implies 3^{2k+2} = 64m + 8k + 9\)

Step 4: Inductive Step (Prove for n=k+1)
We need to prove that \(P(k+1) = 3^{2(k+1)+2} – 8(k+1) – 9\) is divisible by 64.
\(P(k+1) = 3^{2k+2+2} – 8k – 8 – 9\)
\(P(k+1) = 3^2 \cdot 3^{2k+2} – 8k – 17\)
\(P(k+1) = 9 \cdot (3^{2k+2}) – 8k – 17\)

Substitute \(3^{2k+2}\) from Step 3:
\(P(k+1) = 9 \cdot (64m + 8k + 9) – 8k – 17\)
\(P(k+1) = 9(64m) + 72k + 81 – 8k – 17\)
\(P(k+1) = 9(64m) + (72k – 8k) + (81 – 17)\)
\(P(k+1) = 9(64m) + 64k + 64\)
\(P(k+1) = 64 \cdot (9m + k + 1)\)

Since \(P(k+1)\) is a multiple of 64, the statement is true for \(n=k+1\).
Therefore, the expression is divisible by 64 for all \(n \in N\).

Let \(P(n)\) be the statement \(2^{n+4}+12 \ge k(n+4)\). We need to find the greatest integer \(k\) that satisfies this for all \(n \in N\).

Let’s test the first few values of \(n\).

Case n=1:
\(2^{1+4} + 12 \ge k(1+4)\)
\(2^5 + 12 \ge 5k\)
\(32 + 12 \ge 5k\)
\(44 \ge 5k \implies k \le \frac{44}{5} \implies k \le 8.8\)

Case n=2:
\(2^{2+4} + 12 \ge k(2+4)\)
\(2^6 + 12 \ge 6k\)
\(64 + 12 \ge 6k\)
\(76 \ge 6k \implies k \le \frac{76}{6} \implies k \le 12.66…\)

Case n=3:
\(2^{3+4} + 12 \ge k(3+4)\)
\(2^7 + 12 \ge 7k\)
\(128 + 12 \ge 7k\)
\(140 \ge 7k \implies k \le \frac{140}{7} \implies k \le 20\)

The condition must hold for *all* \(n \in N\). The check for \(n=1\) gives the most restrictive condition so far (\(k \le 8.8\)). The greatest integer \(k\) that satisfies \(k \le 8.8\) is \(k=8\).

Let’s check if \(k=8\) works for all \(n\). We need to prove \(2^{n+4}+12 \ge 8(n+4)\) for all \(n \in N\).

Step 1: Base Case (n=1)
\(2^5 + 12 \ge 8(5) \implies 32 + 12 \ge 40 \implies 44 \ge 40\). (True)

Step 2: Inductive Hypothesis
Assume \(P(m)\) is true: \(2^{m+4}+12 \ge 8(m+4)\) for some \(m \in N\).

Step 3: Inductive Step (Prove for n=m+1)
We need to prove \(P(m+1)\): \(2^{(m+1)+4}+12 \ge 8((m+1)+4) \implies 2^{m+5}+12 \ge 8(m+5)\).
\(2^{m+5}+12 = 2 \cdot 2^{m+4} + 12\)
From the hypothesis, \(2^{m+4} \ge 8(m+4) – 12\).
\(2 \cdot 2^{m+4} + 12 \ge 2 \cdot (8m + 32 – 12) + 12\)
\(2 \cdot 2^{m+4} + 12 \ge 2 \cdot (8m + 20) + 12\)
\(2^{m+5} + 12 \ge 16m + 40 + 12 = 16m + 52\).

We need to show \(16m + 52 \ge 8(m+5)\), which is \(16m + 52 \ge 8m + 40\).
\(16m – 8m \ge 40 – 52\)
\(8m \ge -12\). This is true for all \(m \in N\).

Since the inductive step is true, \(k=8\) satisfies the inequation for all \(n \in N\).
The greatest positive integer \(k\) is 8.

Let \(P(n) = 5^{2n} – 48n + k\). The statement \(P(n)\) must be true for all \(n \in N\), so it must be true for \(n=1\).

Step 1: Test for n=1
\(P(1) = 5^{2(1)} – 48(1) + k\)
\(P(1) = 25 – 48 + k = -23 + k\).

Step 2: Find k
For \(P(1)\) to be divisible by 24, \(-23 + k\) must be a multiple of 24.
Let \(-23 + k = 24m\) for some integer \(m\).

If \(m = 0\), \(k = 23\).
If \(m = 1\), \(k = 23 + 24 = 47\).
If \(m = -1\), \(k = 23 – 24 = -1\).

The question asks for the least positive integral value of \(k\).
The least positive integer in the set \(\{… -1, 23, 47, …\}\) is 23.

Step 3: Verification (Optional but recommended)
Let’s check if \(k=23\) works for \(n=2\).
\(P(2) = 5^{2(2)} – 48(2) + 23\)
\(P(2) = 5^4 – 96 + 23 = 625 – 96 + 23 = 529 + 23 = 552\).
\(552 \div 24 = 23\).
Since \(P(2)\) is also divisible by 24, \(k=23\) is the correct value.

Let \(P(n) = 2^{3n} – 7n – 1\).

Method 1: Testing \(n=1\) and \(n=2\)

Case n=1:
\(P(1) = 2^{3(1)} – 7(1) – 1 = 2^3 – 7 – 1 = 8 – 7 – 1 = 0\).
Since 0 is divisible by any non-zero integer, this case doesn’t help narrow it down.

Case n=2:
\(P(2) = 2^{3(2)} – 7(2) – 1 = 2^6 – 14 – 1 = 64 – 15 = 49\).
49 is divisible by 49. Thus, 49 is the only possible answer.

Method 2: Binomial Theorem (Proof)

\(P(n) = 2^{3n} – 7n – 1 = (2^3)^n – 7n – 1 = 8^n – 7n – 1\)
We can write \(8^n\) as \((1 + 7)^n\).
Using the Binomial Theorem, \((1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + … + \binom{n}{n}x^n\).
\(P(n) = (1 + 7)^n – 7n – 1\)
\(P(n) = \left[ \binom{n}{0} + \binom{n}{1}7 + \binom{n}{2}7^2 + \binom{n}{3}7^3 + … \right] – 7n – 1\)
\(P(n) = \left[ 1 + (n)(7) + \frac{n(n-1)}{2}(49) + … \right] – 7n – 1\)
\(P(n) = (1 + 7n) – (7n + 1) + 49 \cdot [\text{… terms …}] = 49 \cdot [\text{… terms …}]\)
Therefore, \(P(n)\) is always divisible by 49.

Let \(P(n) = 3 \cdot 5^{2n+1} + 2^{3n+1}\).

Step 1: Base Case (n=1)
\(P(1) = 3 \cdot 5^{2(1)+1} + 2^{3(1)+1} = 3 \cdot 5^3 + 2^4\)
\(P(1) = 3 \cdot 125 + 16 = 375 + 16 = 391\).

Step 2: Check Divisibility
Let’s check the options for 391:
\(391 \div 17 = 23\). (Divisible)
\(391 \div 23 = 17\). (Divisible)
Let’s check \(n=2\).

Step 3: Base Case (n=2)
\(P(2) = 3 \cdot 5^{2(2)+1} + 2^{3(2)+1} = 3 \cdot 5^5 + 2^7\)
\(P(2) = 3 \cdot 3125 + 128 = 9375 + 128 = 9503\).
\(9503 \div 17 = 559\). (Divisible)
\(9503 \div 23 = 413.17…\) (Not divisible)

The only common divisor for all \(n \in N\) is 17.

Let \(P(n) = 49^n + 16n + k\). The statement must be true for \(n=1\).

Step 1: Base Case (n=1)
\(P(1) = 49^1 + 16(1) + k = 49 + 16 + k = 65 + k\).

Step 2: Find k
For \(P(1)\) to be divisible by 64, \(65 + k\) must be a multiple of 64.
Let \(65 + k = 64m\) for some integer \(m\).

If \(m=1\), \(65+k = 64 \implies k = -1\).
If \(m=2\), \(65+k = 128 \implies k = 63\).
The possible integer values for \(k\) are \(\{… -65, -1, 63, 127, …\}\)
The question asks for the least positive integer \(k\). From the set, the least positive integer is 63.

Step 3: Proof (Binomial Theorem for k=-1)
Let’s prove \(49^n + 16n – 1\) is divisible by 64.
\(P(n) = 49^n + 16n – 1 = (48+1)^n + 16n – 1\)
\(P(n) = \left[ \binom{n}{0}48^0 + \binom{n}{1}48^1 + \binom{n}{2}48^2 + … \right] + 16n – 1\)
\(P(n) = \left[ 1 + 48n + \frac{n(n-1)}{2}(2304) + … \right] + 16n – 1\)
\(P(n) = (1 + 48n + 16n – 1) + \left( \frac{n(n-1)}{2} \cdot 2304 + … \right)\)
\(P(n) = 64n + (\text{terms divisible by 64, since } 2304 = 64 \times 36)\)
\(P(n) = 64n + 64 \cdot [\text{… terms …}]\)
Since \(49^n + 16n – 1\) is divisible by 64, then \(P(n) = (49^n + 16n – 1) + 64\) is also divisible by 64. This corresponds to \(k=63\).

Let \(P(n) = 3^{2n+1} + 2^{n+2}\).

Step 1: Base Case (n=1)
\(P(1) = 3^{2(1)+1} + 2^{1+2} = 3^3 + 2^3 = 27 + 8 = 35\).
35 is divisible by 5 and 7.

Step 2: Base Case (n=2)
\(P(2) = 3^{2(2)+1} + 2^{2+2} = 3^5 + 2^4 = 243 + 16 = 259\).
\(259 \div 7 = 37\). (Divisible by 7)
\(259 \div 5\) is not an integer.
The common divisor is 7.

Let \(P(n) = 2 \cdot 4^{2n+1} + 3^{3n+1}\).

Step 1: Base Case (n=1)
\(P(1) = 2 \cdot 4^{2(1)+1} + 3^{3(1)+1} = 2 \cdot 4^3 + 3^4\)
\(P(1) = 2 \cdot 64 + 81 = 128 + 81 = 209\).

Step 2: Check Divisibility
$209 \div 11 = 19$.
The answer is 11.

Step 3: Proof (Modular Arithmetic)
\(P(n) = 2 \cdot 4 \cdot 4^{2n} + 3 \cdot 3^{3n} = 8 \cdot (16)^n + 3 \cdot (27)^n\)
We check divisibility by 11 (modulo 11):
\(16 \equiv 5 \pmod{11}\)
\(27 \equiv 5 \pmod{11}\)
So, \(P(n) \equiv 8 \cdot (5)^n + 3 \cdot (5)^n \pmod{11}\)
\(P(n) \equiv (8 + 3) \cdot 5^n \pmod{11}\)
\(P(n) \equiv 11 \cdot 5^n \pmod{11} \equiv 0 \pmod{11}\)
Therefore, the expression is divisible by 11.

Let \(P(n) = 7^{2n} + 2^{3n-3} \cdot 3^{n-1}\). This is valid for \(n \ge 1\).

Step 1: Simplify the expression
\(P(n) = (7^2)^n + \frac{2^{3n}}{2^3} \cdot \frac{3^n}{3^1} = 49^n + \frac{(2^3)^n \cdot 3^n}{8 \cdot 3}\)
\(P(n) = 49^n + \frac{8^n \cdot 3^n}{24} = 49^n + \frac{(8 \cdot 3)^n}{24} = 49^n + \frac{24^n}{24} = 49^n + 24^{n-1}\).

Step 2: Base Case (n=1)
\(P(1) = 49^1 + 24^{1-1} = 49 + 24^0 = 49 + 1 = 50\).
50 is divisible by 25.

Step 3: Base Case (n=2)
\(P(2) = 49^2 + 24^{2-1} = 2401 + 24 = 2425\).
\(2425 \div 25 = 97\).
Therefore, the expression is divisible by 25.

This is a repeat question.

Let \(P(n) = 3 \cdot 5^{2n+1} + 2^{3n+1}\).

Step 1: Base Case (n=1)
\(P(1) = 3 \cdot 5^{3} + 2^{4} = 3 \cdot 125 + 16 = 375 + 16 = 391\).

Step 2: Check Divisibility
\(391 \div 17 = 23\). The answer is 17.

This is a repeat question.

Let \(P(n) = 2^{3n} – 7n – 1\).

Method: Binomial Theorem

\(P(n) = (2^3)^n – 7n – 1 = 8^n – 7n – 1\)
\(P(n) = (1 + 7)^n – 7n – 1\)
\(P(n) = \left[ 1 + \binom{n}{1}7 + \binom{n}{2}7^2 + … \right] – 7n – 1\)
\(P(n) = (1 + 7n + 49 \cdot [\text{… terms …}]) – 7n – 1 = 49 \cdot [\text{… terms …}]\)
The expression is always divisible by 49.

Let \(P(n) = n(n+1)(n+2)(n+3)\).

Method 1: Base Case (n=1)
\(P(1) = 1(2)(3)(4) = 24\).
Since 24 is divisible by 24, this is the largest possible option.

Method 2: Logical Argument
\(P(n)\) is the product of four consecutive integers.
Among any four consecutive integers:

1. There is exactly one multiple of 4.
2. There is at least one other multiple of 2.
3. There is at least one multiple of 3.

The product must be divisible by \(4 \times 2 \times 3 = 24\).

Method 3: Combinatorial Argument
\(\frac{P(n)}{24} = \frac{n(n+1)(n+2)(n+3)}{4 \times 3 \times 2 \times 1} = \binom{n+3}{4}\).
Since \(\binom{n+3}{4}\) is always an integer, \(P(n)\) must be divisible by 24.

Let \(P(n) = 10^n + 3 \times 4^{n+2} + 5\).

Step 1: Base Case (n=1)
\(P(1) = 10^1 + 3 \times 4^{1+2} + 5 = 10 + 3 \times 4^3 + 5\)
\(P(1) = 10 + 3 \times 64 + 5 = 10 + 192 + 5 = 207\).

Step 2: Check Divisibility
The sum of the digits of 207 is \(2+0+7 = 9\). Since the sum of digits is divisible by 9, 207 is divisible by 9.
\(207 \div 9 = 23\).

This is a repeat question.

Let \(P(n) = 3^{2n+2} – 8n – 9\).

Method: Binomial Theorem

\(P(n) = 3^2 \cdot (3^2)^n – 8n – 9 = 9 \cdot 9^n – 8n – 9\)
\(P(n) = 9 \cdot (1 + 8)^n – 8n – 9\)
\(P(n) = 9 \cdot \left[ 1 + n(8) + \frac{n(n-1)}{2}(8^2) + … \right] – 8n – 9\)
\(P(n) = (9 + 72n + \text{terms divisible by 64}) – 8n – 9\)
\(P(n) = (72n – 8n) + (9 – 9) + (\text{multiple of 64})\)
\(P(n) = 64n + (\text{multiple of 64})\)
The expression is divisible by 64.

This is a repeat question.

Let \(P(n) = 7^{2n} + 3^{n-1} \cdot 2^{3n-3}\) (for \(n \ge 1\)).

Step 1: Simplify the expression
\(P(n) = (7^2)^n + \frac{3^n}{3} \cdot \frac{2^{3n}}{2^3} = 49^n + \frac{3^n \cdot 8^n}{24} = 49^n + \frac{24^n}{24} = 49^n + 24^{n-1}\).

Step 2: Base Case (n=1)
\(P(1) = 49^1 + 24^{1-1} = 49 + 24^0 = 49 + 1 = 50\).
50 is divisible by 25. The answer is 25.

This is a repeat question.

Let \(P(n) = 10^n + 3 \times 4^{n+2} + 5\).

Step 1: Base Case (n=1)
\(P(1) = 10^1 + 3 \times 4^{1+2} + 5 = 10 + 3 \times 4^3 + 5\)
\(P(1) = 10 + 192 + 5 = 207\).

Step 2: Check Divisibility
The sum of the digits of 207 is \(2+0+7 = 9\), so it is divisible by 9.

This is a repeat question.

Let \(P(n) = 3 \cdot 5^{2n+1} + 2^{3n+1}\).

Step 1: Base Case (n=1)
\(P(1) = 3 \cdot 5^{3} + 2^{4} = 3 \cdot 125 + 16 = 375 + 16 = 391\).

Step 2: Check Divisibility
\(391 \div 17 = 23\). The answer is 17.

This is a repeat question.

Let \(P(n) = 2 \cdot 4^{2n+1} + 3^{3n+1}\).

Step 1: Base Case (n=1)
\(P(1) = 2 \cdot 4^{3} + 3^{4} = 2 \cdot 64 + 81 = 128 + 81 = 209\).

Step 2: Check Divisibility
\(209 \div 11 = 19\). The answer is 11.

Let \(P(n)\) be the statement \(1^2+2^2+…+n^2 > \frac{n^3}{3}\).

Step 1: Base Case (n=1)
LHS = \(1^2 = 1\).
RHS = \(\frac{1^3}{3} = \frac{1}{3}\).
Since \(1 > \frac{1}{3}\), \(P(1)\) is true.

Step 2: Inductive Hypothesis
Assume \(P(k)\) is true for some \(k \in N\): \(1^2+2^2+…+k^2 > \frac{k^3}{3}\)

Step 3: Inductive Step (Prove for n=k+1)
We need to prove \(P(k+1)\): \(1^2+2^2+…+k^2+(k+1)^2 > \frac{(k+1)^3}{3}\).
LHS = \((1^2+2^2+…+k^2) + (k+1)^2\)
From our hypothesis, \(LHS > \frac{k^3}{3} + (k+1)^2\)
\(LHS > \frac{k^3}{3} + k^2 + 2k + 1 = \frac{k^3 + 3k^2 + 6k + 3}{3}\).

The RHS we need to prove is: \(\frac{(k+1)^3}{3} = \frac{k^3 + 3k^2 + 3k + 1}{3}\).
We must show: \(\frac{k^3 + 3k^2 + 6k + 3}{3} > \frac{k^3 + 3k^2 + 3k + 1}{3}\).
Comparing numerators: \(k^3 + 3k^2 + 6k + 3 > k^3 + 3k^2 + 3k + 1\)
\(6k + 3 > 3k + 1 \implies 3k > -2\).
Since \(k \in N\), \(k \ge 1\), this is always true. Thus, \(P(k+1)\) is true.

This is a repeat question.

Let \(P(n) = 3^{2n+2} – 8n – 9\).

Method: Binomial Theorem

\(P(n) = 9 \cdot 9^n – 8n – 9 = 9 \cdot (1 + 8)^n – 8n – 9\)
\(P(n) = 9 \cdot \left[ 1 + n(8) + \frac{n(n-1)}{2}(8^2) + … \right] – 8n – 9\)
\(P(n) = (9 + 72n + \text{terms divisible by 64}) – 8n – 9\)
\(P(n) = 64n + (\text{multiple of 64})\)
The expression is divisible by 64.

This is a repeat question. We need to find the *largest* integer \(\lambda\).

Let \(P(n) = 2^{3n} – 7n – 1\).

Step 1: Base Case (n=1)
\(P(1) = 2^3 – 7(1) – 1 = 8 – 7 – 1 = 0\).

Step 2: Base Case (n=2)
\(P(2) = 2^6 – 7(2) – 1 = 64 – 14 – 1 = 49\).
This suggests \(\lambda\) could be 49.

Step 3: Base Case (n=3)
\(P(3) = 2^9 – 7(3) – 1 = 512 – 21 – 1 = 490\).
\(490 = 10 \times 49\). This is also divisible by 49.

The greatest common divisor for \(P(2)\) and \(P(3)\) is 49. Therefore, the largest integer value for \(\lambda\) is 49.

Let \(P(n) = n(n+1)(n+2)\).

Method 1: Base Cases
\(P(1) = 1(2)(3) = 6\). (Divisible by 6)
\(P(2) = 2(3)(4) = 24\). (Divisible by 6, 8, 12)
\(P(3) = 3(4)(5) = 60\). (Divisible by 6, 12, but not 8)
Since \(P(3)=60\) is not divisible by 8, options 2 and 3 are incorrect. Since \(P(1)=6\) is not divisible by 12, option 4 is incorrect. The only divisor that works for all cases is 6.

Method 2: Logical Argument
\(P(n)\) is the product of three consecutive integers.
In any set of three consecutive integers, there must be at least one multiple of 3 and at least one multiple of 2. Since 2 and 3 are coprime, the product must be divisible by \(2 \times 3 = 6\).

This is a repeat question.

Let \(P(n) = 3 \cdot 5^{2n+1} + 2^{3n+1}\).

Step 1: Base Case (n=1)
\(P(1) = 3 \cdot 5^{3} + 2^{4} = 3 \cdot 125 + 16 = 375 + 16 = 391\).

Step 2: Check Divisibility
\(391 \div 17 = 23\). The answer is 17.

Let \(P(n) = 49^n + 16n – 1\).

Method 1: Base Cases
\(P(1) = 49^1 + 16(1) – 1 = 49 + 16 – 1 = 64\). (Divisible by 64)
\(P(2) = 49^2 + 16(2) – 1 = 2401 + 32 – 1 = 2432\).
\(2432 \div 64 = 38\). (Divisible by 64).
The answer is 64.

Method 2: Binomial Theorem
\(P(n) = (48+1)^n + 16n – 1\)
\(P(n) = \left[ 1 + \binom{n}{1}48 + \binom{n}{2}48^2 + … \right] + 16n – 1\)
\(P(n) = (1 + 48n) + 16n – 1 + (\text{terms divisible by } 48^2 = 2304)\)
Since \(2304 = 64 \times 36\), all terms from \(\binom{n}{2}\) onwards are divisible by 64.
\(P(n) = (1 + 48n + 16n – 1) + (\text{multiple of 64})\)
\(P(n) = 64n + (\text{multiple of 64})\)
The expression is divisible by 64.

Let \(P(n) = 10^n + 3 \times 4^{n+2} + \lambda\). The statement must be true for \(n=1\).

Step 1: Base Case (n=1)
\(P(1) = 10^1 + 3 \times 4^{1+2} + \lambda = 10 + 3 \times 4^3 + \lambda\)
\(P(1) = 10 + 3 \times 64 + \lambda = 10 + 192 + \lambda = 202 + \lambda\).

Step 2: Find \(\lambda\)
For \(P(1)\) to be divisible by 9, \(202 + \lambda\) must be a multiple of 9.
The sum of digits of 202 is \(2+0+2=4\). So, \(202 \equiv 4 \pmod{9}\).
We need \(4 + \lambda \equiv 0 \pmod{9}\).
This means \(4 + \lambda\) must be a multiple of 9.
The least positive integer \(\lambda\) that satisfies this is \(\lambda = 5\), since \(4+5=9\).

This is a repeat question. We need to find the *largest* integer \(k\).

Let \(P(n) = 3^{2n+2} – 8n – 9\).

Step 1: Base Case (n=1)
\(P(1) = 3^{2(1)+2} – 8(1) – 9 = 3^4 – 8 – 9 = 81 – 17 = 64\).
The largest possible value for \(k\) from the options is 64.

Step 2: Base Case (n=2)
\(P(2) = 3^{2(2)+2} – 8(2) – 9 = 3^6 – 16 – 9 = 729 – 25 = 704\).
\(704 \div 64 = 11\).

Since the HCF of \(P(1)\) and \(P(2)\) is 64, the largest such integer \(k\) is 64.