AP EAMCET · Maths · Binomial Theorem
If \(a_n=\sum_{r=0}^n \frac{1}{{ }^n C_r}\) then \(\sum_{r=0}^n \frac{r}{{ }^n C_r}=\)
- A \((\mathrm{n}-1) \mathrm{a}\)
- B n. an
- C \(\frac{n}{2} \cdot a_n\)
- D \(\mathrm{an}^{+}{ }_1\)
Answer & Solution
Correct Answer
(C) \(\frac{n}{2} \cdot a_n\)
Step-by-step Solution
Detailed explanation
Let \(b=\sum_{r=0}^n \frac{r}{{ }^n C_r}\) ....(i) Replace \(r\) by \(n-r\) \(b=\sum_{r=0}^n \frac{n-r}{{ }^n C_{n-r}}\) ....(ii) Adding (i) and (ii)…
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