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  • Let n be  an odd  natural number  and  <img alt="A=\sum_{r=1}^{\frac{n-1}{2}}

Let n be  an odd  natural number  and  A=\sum_{r=1}^{\frac{n-1}{2}} \frac{1}{{ }^n C_r}. Then value  of \sum_{r=1}^n \frac{r}{{ }^n C_r} is  equal to

Option: 1

 n(A – 1)


Option: 2

n( A + 1)


Option: 3

\frac{\mathrm{nA}}{2}


Option: 4

 nA


Answers (1)

best_answer

\sum_{r=1}^n \frac{r}{{ }^n C_r}=\sum_{r=0}^{n-1} \frac{(n-r)}{{ }^n C_r}

=\sum_{r=0}^{n-1} \frac{n}{{ }^n C_r}-\sum_{r=0}^{n-1} \frac{r}{{ }^n C_r}

\sum_{r=1}^n \frac{r}{{ }^n C_r}=n \sum_{r=0}^{n-1} \frac{1}{{ }^n C_r}+n+n=n 2 A+2 n

=\sum_{r=1}^n \frac{r}{n C_r}=n(A+1)

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chirag

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