Let C_r denote the coefficient of x^r in the binomial expansion of (1+x)^n, n N, 0 r n.

Question

Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1+x)^n$, $n\in N$, $0\le r\le n$. If

$$P_n = C_0 - \frac{2}{2}C_1 + \frac{2^2}{3}C_2 - \frac{2^3}{4}C_3 + \cdots + \frac{(-2)^n}{n+1}C_n$$

then the value of $\displaystyle\sum_{n=1}^{25}\frac{1}{P_{2n}}$ equals

Accepted Answer

$$P_n=\sum_{r=0}^n \frac{(-2)^r}{r+1}\binom{n}{r} = \frac{1}{n+1}\sum_{r=0}^n \binom{n+1}{r+1}(-2)^r = \frac{-1}{2(n+1)}\sum_{r=0}^n \binom{n+1}{r+1}(-2)^{r+1}=\frac{-1}{2(n+1)}\big[(1-2)^{n+1}-1\big]$$

$$P_n = \frac{1}{2(n+1)}\big[1-(-1)^{n+1}\big]$$

For $n=2m$: $P_{2m}=\dfrac{1}{2(2m+1)}\big[1-(-1)^{2m+1}\big]=\dfrac{1}{2m+1}$.

So $\dfrac{1}{P_{2n}}=2n+1$, and

$$\sum_{n=1}^{25}\frac{1}{P_{2n}}=\sum_{n=1}^{25}(2n+1)=3+5+\cdots+51=\frac{25}{2}(3+51)=25\times27=675$$

Solutions · 1