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Question:
Answered & Verified by Expert
If $C_j$ stands for ${ }^n C_j$, then
$$
\frac{\mathrm{C}_1}{\mathrm{C}_0}+\frac{2 \times \mathrm{C}_2}{\mathrm{C}_1}+\frac{3 \times \mathrm{C}_3}{\mathrm{C}_2}+\ldots+\frac{\mathrm{n} \times \mathrm{C}_{\mathrm{n}}}{\mathrm{C}_{\mathrm{n}-1}}=
$$
Options:
$$
\frac{\mathrm{C}_1}{\mathrm{C}_0}+\frac{2 \times \mathrm{C}_2}{\mathrm{C}_1}+\frac{3 \times \mathrm{C}_3}{\mathrm{C}_2}+\ldots+\frac{\mathrm{n} \times \mathrm{C}_{\mathrm{n}}}{\mathrm{C}_{\mathrm{n}-1}}=
$$
Solution:
2386 Upvotes
Verified Answer
The correct answer is:
$\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}$
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
$$
\begin{aligned}
& \frac{C_1}{C_0}+\frac{2 C_2}{C_1}+\ldots+\frac{n C_n}{C_{n-1}}=n+(n-1)+(n-2)+\ldots \\
& =n+(n-1)+(n-2)+\ldots+3+2+1 \\
& =\sum_{k=1}^n k
\end{aligned}
$$
$$
\frac{r \cdot{ }^n C_r}{{ }^n C_{r-1}}=\frac{r \cdot \frac{n !}{r !(n-r) !}}{\frac{n !}{(r-1) !(n-r+1) !}}=(n-r+1)
$$
Hence
$$
\begin{aligned}
& \frac{C_1}{C_0}+\frac{2 C_2}{C_1}+\ldots+\frac{n C_n}{C_{n-1}}=n+(n-1)+(n-2)+\ldots \\
& =n+(n-1)+(n-2)+\ldots+3+2+1 \\
& =\sum_{k=1}^n k
\end{aligned}
$$
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