Question

If $(1+x{)}^{15}=a_0+a_{1}x+...+a_{15}{x}^{15}$, then $\sum _{r=1}^{15}r\left(\frac{a_r}{a_{r-1}}\right)$ is

• A. $110$
• B. $115$
• C. $120$
• D. $135$

Answer 1

The correct option is C

$120$

Explanation for the correct option:

Step 1: Expand $(1+x{)}^{15}$

${(1+x)}^{15}=\left(\begin{array}{c}15\\ 0\end{array}\right)+\left(\begin{array}{c}15\\ 1\end{array}\right)x+\left(\begin{array}{c}15\\ 2\end{array}\right)x^2+...+\left(\begin{array}{c}15\\ 15\end{array}\right)x^{15}$

Step 2: Comparing equation in step 1 with the given equation.

$a_0=\left(\begin{array}{c}15\\ 0\end{array}\right),a_1=\left(\begin{array}{c}15\\ 1\end{array}\right),a_2=\left(\begin{array}{c}15\\ 2\end{array}\right),.....,a_{15}=\left(\begin{array}{c}15\\ 15\end{array}\right)$

$\therefore$$a_r=\left(\begin{array}{c}15\\ r\end{array}\right)$

Step 3: Calculate $\begin{array}{l}\mathbf{}\frac{\mathbf{r}{\mathbf{a}}_{\mathbf{r}}}{{\mathbf{a}}_{\mathbf{r}\mathbf{-}\mathbf{1}}}\begin{array}{}\end{array}\begin{array}{}\end{array}\end{array}$

$\begin{array}{rcl}\frac{r{a}_r}{a_{r-1}}& =& \frac{r\left(\begin{array}{c}15\\ r\end{array}\right)}{\left(\begin{array}{c}15\\ r-1\end{array}\right)}\\ & =& \frac{{\displaystyle \frac{r(15!)}{(15-r)!r!}}}{{\displaystyle \frac{15!}{\{15-(r-1\left)\right\}!(r-1)!}}}\\ & =& \frac{r(15-r+1)}{r}\\ & =& 15-r+1\\ & =& 16-r\end{array}$

Step 4: Calculate $\begin{array}{l}\sum _{r=1}^{15}\frac{r{a}_r}{a_{r-1}}\begin{array}{}\end{array}\begin{array}{}\end{array}\end{array}$

$\begin{array}{rcl}\sum _{r=1}^{15}\frac{r{a}_r}{a_{r-1}}=\sum _{r=1}^{15}(16-r)\begin{array}{}\end{array}& =& 15+14+13+...+1\begin{array}{}\end{array}\\ & =& 120\end{array}$

$\mathbf{\therefore }\mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{1}}^{\mathbf{15}}\frac{\mathbf{r}{\mathbf{a}}_{\mathbf{r}}}{{\mathbf{a}}_{\mathbf{r}\mathbf{-}\mathbf{1}}}\mathbf{=}\mathbf{120}\begin{array}{}\end{array}$

Hence, option $\mathbf{\left(}\mathit{C}\mathbf{\right)}$ is the correct option.