Question

Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1+x)^n$, $n\in\mathbb{N$, $0\le r\le n$. 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.
\[ \sum_{n=1}^{25} \frac{1}{2n}P_n = \sum_{n=1}^{25} \frac{1}{2n}\cdot \frac{1-(-1)^{n+1}}{2(n+1)} \] \[ = \sum_{\substack{n=1
n\text{ odd}}}^{25} \frac{1}{2n}\cdot\frac{2}{2(n+1)} \] \[ = \sum_{\substack{n=1
n\text{ odd}}}^{25} \frac{1}{2n(n+1)} \]
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: $\boxed{675}$