Question

Let $\lambda$ is the coefficient of $x^2$ in the expansion $(1+x)(1-3x)(1+5x)(1-7x)...(1-23x)(1+25x)$ then number of positive divisors of $|\lambda|$ is

• A. 4
• B. 6
• C. 8
• D. 10

Answer 1

Answer: (C)

8

Answer 2

The given expression is the product of 13 terms: $P(x) = (1+x)(1-3x)(1+5x)(1-7x)...(1-23x)(1+25x)$

This can be written as $P(x) = \prod_{k=1}^{13} (1+a_k x)$, where $a_k = (-1)^{k+1}(2k-1)$. The coefficients $a_k$ are $1, -3, 5, -7, 9, -11, 13, -15, 17, -19, 21, -23, 25$.

We want to find the coefficient of $x^2$ in the expansion of $P(x)$. The expansion of a product of terms $(1+a_k x)$ is given by: $\prod_{k=1}^n (1+a_k x) = 1 + (\sum_{k=1}^n a_k) x + (\sum_{1 \le i < j \le n} a_i a_j) x^2 + O(x^3)$. In this case, $n=13$. The coefficient of $x^2$ is $\lambda = \sum_{1 \le i < j \le 13} a_i a_j$.

We know the identity $(\sum_{k=1}^{13} a_k)^2 = \sum_{k=1}^{13} a_k^2 + 2 \sum_{1 \le i < j \le 13} a_i a_j$. So, $\lambda = \frac{1}{2} \left[ \left( \sum_{k=1}^{13} a_k \right)^2 - \sum_{k=1}^{13} a_k^2 \right]$.

Let $S = \sum_{k=1}^{13} a_k$. $S = 1 + (-3) + 5 + (-7) + ... + (-23) + 25$ $S = (1-3) + (5-7) + (9-11) + (13-15) + (17-19) + (21-23) + 25$ $S = (-2) + (-2) + (-2) + (-2) + (-2) + (-2) + 25$ $S = 6 \times (-2) + 25 = -12 + 25 = 13$.

Let $S_2 = \sum_{k=1}^{13} a_k^2$. $a_k^2 = ((-1)^{k+1}(2k-1))^2 = (2k-1)^2$. $S_2 = \sum_{k=1}^{13} (2k-1)^2 = 1^2 + 3^2 + 5^2 + ... + 25^2$. This is the sum of the squares of the first 13 odd numbers. The formula for the sum of squares of the first $n$ odd numbers is $\frac{n(2n-1)(2n+1)}{3}$. For $n=13$: $S_2 = \frac{13(2 \times 13 - 1)(2 \times 13 + 1)}{3} = \frac{13(25)(27)}{3} = 13 \times 25 \times 9 = 13 \times 225 = 2925$.

Now we can calculate $\lambda$: $\lambda = \frac{1}{2} (S^2 - S_2) = \frac{1}{2} (13^2 - 2925) = \frac{1}{2} (169 - 2925) = \frac{1}{2} (-2756) = -1378$.

We need to find the number of positive divisors of $|\lambda|$. $|\lambda| = |-1378| = 1378$. To find the number of positive divisors of 1378, we find its prime factorization. $1378 = 2 \times 689$. To factor 689, we test for divisibility by prime numbers. $\sqrt{689} \approx 26.2$. We check primes up to 23: 2, 3, 5, 7, 11, 13, 17, 19, 23. 689 is not divisible by 2, 3, 5, 7, 11. $689 \div 13 = 53$. So, $1378 = 2 \times 13 \times 53$. The prime factors are 2, 13, and 53, each with an exponent of 1. The number of positive divisors is given by the product of (exponent + 1) for each prime factor. Number of divisors = $(1+1) \times (1+1) \times (1+1) = 2 \times 2 \times 2 = 8$.

The number of positive divisors of $|\lambda|$ is 8.