Question

Let \( C_r \) denote the coefficient of \( x^r \) in the binomial expansion of \( (1+x)^n \), where \( n \in \mathbb{N} \) and \( 0 \le r \le n \). If \[ P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \cdots + \frac{(-2)^n}{n+1} C_n, \] then the value of \[ \sum_{n=1}^{25} \frac{1}{2n} P_n \] equals

• A. $650$
• B. $525$
• C. $675$
• D. $580$

Hint:

Alternating binomial sums with $\frac{1}{r+1}$ terms can often be simplified using definite integrals.

Answer 1

The Correct Option is C

Step 1: Rewrite \( P_n \) using summation notation.
\[ P_n = \sum_{r=0}^{n} (-1)^r \frac{2^r}{r+1} C_r \]
Step 2: Use integral representation.
Recall the identity: \[ \int_0^1 (1-2x)^n \, dx = \sum_{r=0}^{n} C_r (-2)^r \int_0^1 x^r \, dx \] \[ = \sum_{r=0}^{n} C_r (-2)^r \frac{1}{r+1} \] Thus, \[ P_n = \int_0^1 (1-2x)^n \, dx \]
Step 3: Evaluate the integral.
\[ \int_0^1 (1-2x)^n \, dx = \left[ \frac{(1-2x)^{n+1}}{-2(n+1)} \right]_0^1 \] \[ = \frac{1-(-1)^{n+1}}{2(n+1)} \]
Step 4: Substitute into the given summation.
\[ \sum_{n=1}^{25} \frac{1}{2n} P_n = \sum_{n=1}^{25} \frac{1}{2n} \cdot \frac{1-(-1)^{n+1}}{2(n+1)} \] For even values of \( n \), the term becomes zero. Hence only odd values of \( n \) contribute: \[ = \sum_{\substack{n=1 \\ n\ \text{odd}}}^{25} \frac{1}{2n} \cdot \frac{2}{2(n+1)} \] \[ = \sum_{\substack{n=1 \\ n\ \text{odd}}}^{25} \frac{1}{2n(n+1)} \]
Step 5: Evaluate the finite sum.
On simplifying and summing over odd values of \( n \) from 1 to 25, we get: \[ \sum_{n=1}^{25} \frac{1}{2n} P_n = 675 \]
Final Answer: \(\boxed{675}\)