If $C_j$ stands for ${ }^n C_j$, then $$ \frac{\mathrm{C}_1}{\mathrm{C}_0}+\frac{2 \times…
If $C_j$ stands for ${ }^n C_j$, then
$$
\frac{\mathrm{C}_1}{\mathrm{C}_0}+\frac{2 \times \mathrm{C}_2}{\mathrm{C}_1}+\frac{3 \times \mathrm{C}_3}{\mathrm{C}_2}+\ldots+\frac{\mathrm{n} \times \mathrm{C}_{\mathrm{n}}}{\mathrm{C}_{\mathrm{n}-1}}=
$$
- $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^2$
- $\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{\mathrm{k}}{2}$
- $\sum_{\mathrm{k}=1}^{\mathrm{n}} 2 \mathrm{k}$
- $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}$
Solution
General term for given polinomial
$
\frac{r \cdot{ }^n C_r}{{ }^n C_{r-1}}=\frac{r \cdot \frac{n !}{r !(n-r) !}}{\frac{n !}{(r-1) !(n-r+1) !}}=(n-r+1)
$
Hence
$
\begin{aligned}
& \frac{C_1}{C_0}+\frac{2 C_2}{C_1}+\ldots+\frac{n C_n}{C_{n-1}}=n+(n-1)+(n-2)+\ldots \\
& =n+(n-1)+(n-2)+\ldots+3+2+1 \\
& =\sum_{k=1}^n k
\end{aligned}
$
Asked in: AP EAMCET 2023 (15 May Shift 1)
Practice more Methods of Induction and Binomial Theorem questions on Aicharya