JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \( C_{r} \) denote the coefficient of \( x^{r} \) in the binomial expansion of \( (1+x)^{n} \),\(n \in N , 0 \leq r \leq n\). If \( P_{n}=C_{0}-C_{1}+\frac{2^{2}}{3}C_{2}-\frac{2^{3}}{4}C_{3}+....+\frac{(-2)^{n}}{n+1}C_{n}, \) then the value of \( \sum_{n=1}^{25}\frac{1}{P_{2n}} \) equals.
- A 580
- B 525
- C 650
- D 675
Answer & Solution
Correct Answer
(D) 675
Step-by-step Solution
Detailed explanation
\(P_n=\sum_{r=0}^n \frac{{ }^n C_r(-2)^r}{r+1}=\sum_{r=0}^n \frac{1}{(n+1)}{ }^{n+1} C_{r+1}(-2)^r\) \(=\frac{-1}{2(n+1)} \sum_{r=0}^n{ }^{n+1} C_{r+1}(-2)^{r+1}\) \(=\frac{-1}{2( n +1)}\left[(1-2)^{ n +1}-1\right]\) \(P_n=\frac{1}{2(n+1)}\left[1-(-1)^{n+1}\right]\)…
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