Q58 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics
If sum of the coefficients of $x^r(r=0,1,2, \ldots, 2 n)$ in the expansion of $\left(1+3 x-2 x^2\right)^n$ is 128 , then $\sum_{r=1}^{2 n} r \frac{(2 n)_{C_r}}{(2 n)_{C_{r-1}}}=$
- A120
- B135
- C90
- Correct answerD105
Solution
Given sum of coefficient in expansion of $\left(1+3 x-2 x^2\right)^n$ is 128
$\therefore$ For sum of coefficient, putting $n=1$ in the expression $\left(1+3 x-2 x^2\right)^n$
$\Rightarrow \quad(1+3 \cdot 1-2 \cdot 1)^n=128 \Rightarrow 2^n=2^7 \Rightarrow n=7$
Now $\sum_{r=1}^{2 n} \frac{r \cdot{ }^{2 n} C_r}{{ }^{2 n} C_{r-1}}=\sum_{r=1}^{2 n} \frac{r \cdot \frac{2 n !}{r !(2 n-r) !}}{\frac{2 n !}{(r-1) !(2 n-r+1) !}}$
$=\sum_{r=1}^{2 n} \frac{r \cdot(2 n-r+1)}{r}=\sum_{r=1}^{2 n}(2 n-r+1) \quad[\because n=7]$
$=(2 n+1)(2 n)-\frac{2 n(2 n+1)}{2}$
$=(14+1)(14)-7(14+1)=15 \times 7=105$
Hence, the correct answer is option: 'D'.