First slide

Binomial theorem for positive integral Index

Question

If $a = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}$ , then the value of $\sum_{0 \leq i < j \leq n} (\frac{i}{^{n} C_{i}} + \frac{j}{^{n} C_{j}})$ is

Difficult

A

an²
B

$\frac{a^{2} n}{2}$
C

a²n
D

$\frac{n^{2} a}{2}$

Solution

$\begin{matrix} S = \underset{0 \leq i < j \leq n}{\sum \sum} (\frac{i}{^{n} C_{i}} + \frac{j}{^{n} C_{j}}) \\ = \underset{0 \leq i < j \leq n}{\sum \sum} (\frac{n - i}{^{n} C_{n - i}} + \frac{n - j}{^{n} C_{n - j}}) \\ = n \underset{0 \leq i < j \leq n}{\sum \sum} (\frac{1}{^{n} C_{i}} + \frac{1}{^{n} C_{j}}) - S \\ \Rightarrow S = \frac{n}{2} \underset{0 \leq i < j \leq n}{\sum \sum} (\frac{1}{^{n} C_{i}} + \frac{1}{^{n} C_{j}}) \\ = \frac{n}{2} (\sum_{r = 0}^{n - 1} \frac{n - r}{^{n} C_{r}} + \sum_{r = 1}^{n} \frac{r}{^{n} C_{r}}) \\ = \frac{n}{2} (\sum_{r = 0}^{n} \frac{n}{^{n} C_{r}}) = \frac{n^{2} a}{2} \end{matrix}$

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