Question

If $C_0, C_1, C_2, ......, C_n$ be the coefficients in the expansion of $(1 + x)^n$, then $\frac{2^{2}.C_{0}}{1.2} + \frac{2^{3}.C_{1}}{2.3}+ ..... + \frac{2^{n+2}. C_{n}}{\left(n+1\right)\left(n+2\right)}$ is

• A. $ \frac{3^{n+1}-2n-5}{\left(n+1\right)\left(n+2\right)}$
• B. $ \frac{3^{n+2}-2n-5}{\left(n+1\right)\left(n+2\right)}$
• C. $ \frac{3^{n+1}+2n-5}{\left(n+1\right)\left(n+2\right)}$
• D. None of these

Answer 1

Answer: (B)

We have,
$t_{r+1} = \frac{2^{r+2}\,{}^{n}C_{r}}{\left(r+1\right)\left(r+2\right)} = \frac{2^{r+2}}{r+2} . \frac{1}{r+1} ^{n}C_{r}$
$ = \frac{2^{r+2}}{r+2} . \frac{1}{n+1} ^{n+1}C_{r+1}$
$= \frac{2^{r+2}}{r+2} . \left(\frac{1}{r+1} ^{n+1}C_{r+1}\right)$
$= \frac{2^{r+2}}{r+2} . \frac{1}{n+2} ^{n+2}C_{r+2}$
$\left[\because \frac{1}{r+1} ^{n}C_{r} = \frac{1}{r+1} ^{n+1}C_{r+1}\right]$
Putting $r = 0, 1, 2, ......, n$ and adding we get,
The given expression
$= \frac{1}{\left(n+1\right)\left(n+2\right) } \left\{2^{2}. ^{n+2}C_{2} + 2^{3}. ^{n+2}C_{3} +.... + 2^{n+2}. ^{n+2}C_{n+2} \right\}$
$= \frac{1}{\left(n+1\right)\left(n+2\right)} \left\{\left(1+2\right)^{n+2}-^{n+2}C_{0}-2. ^{n+2}C_{1}\right\}$
$= \frac{3^{n+2}-2\left(n+2\right)-1}{\left(n+1\right)\left(n+2\right)} = \frac{3^{n+2}-2n-5}{\left(n+1\right)\left(n+2\right)}$