AP EAMCET · Maths · Binomial Theorem
If \(C_j\) stands for \({ }^n C_j\), then
\[
\frac{\mathrm{C}_1}{\mathrm{C}_0}+\frac{2 \times \mathrm{C}_2}{\mathrm{C}_1}+\frac{3 \times \mathrm{C}_3}{\mathrm{C}_2}+\ldots+\frac{\mathrm{n} \times \mathrm{C}_{\mathrm{n}}}{\mathrm{C}_{\mathrm{n}-1}}=
\]
- A \(\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^2\)
- B \(\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{\mathrm{k}}{2}\)
- C \(\sum_{\mathrm{k}=1}^{\mathrm{n}} 2 \mathrm{k}\)
- D \(\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}\)
Answer & Solution
Correct Answer
(D) \(\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}\)
Step-by-step Solution
Detailed explanation
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…
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