Question

The sum $\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+........$ equals

• A. (1) $\frac{1}{(n-1)(n+1)}$
• B. $\frac{1}{(n+1)(n+2)}$
• C. $\frac{1}{2(n-1)(n+1)}$
• D. Zero

Answer 1

Answer: (B)

$\frac{1}{(n+1)(n+2)}$

Answer 2

The given sum is $S = \frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+........$. Assuming $C_k = \binom{n}{k}$ and the series goes up to $k=n$, the sum can be written as: $S = \sum_{k=0}^{n} (-1)^k \frac{\binom{n}{k}}{k+2}$

We use the integral representation of $\frac{1}{k+2}$: $\frac{1}{k+2} = \int_0^1 x^{k+1} dx$

Substitute this into the sum: $S = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \int_0^1 x^{k+1} dx$

Interchange the summation and integration: $S = \int_0^1 \left( \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^{k+1} \right) dx$ $S = \int_0^1 x \left( \sum_{k=0}^{n} \binom{n}{k} (-x)^k \right) dx$

The sum inside the parenthesis is the binomial expansion of $(1-x)^n$: $\sum_{k=0}^{n} \binom{n}{k} (-x)^k = (1-x)^n$

So, the integral becomes: $S = \int_0^1 x (1-x)^n dx$

This integral can be evaluated using substitution. Let $u = 1-x$, so $du = -dx$. When $x=0$, $u=1$. When $x=1$, $u=0$. Also, $x = 1-u$. $S = \int_1^0 (1-u) u^n (-du)$ $S = \int_0^1 (1-u) u^n du$ $S = \int_0^1 (u^n - u^{n+1}) du$

Now, integrate term by term: $S = \left[ \frac{u^{n+1}}{n+1} - \frac{u^{n+2}}{n+2} \right]_0^1$ $S = \left( \frac{1^{n+1}}{n+1} - \frac{1^{n+2}}{n+2} \right) - (0 - 0)$ $S = \frac{1}{n+1} - \frac{1}{n+2}$

Combine the fractions: $S = \frac{(n+2) - (n+1)}{(n+1)(n+2)}$ $S = \frac{1}{(n+1)(n+2)}$