If a_n = sum_{r=0}^n frac{1}{^nC_r},then sum_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$.Let $b_n = \sum_{r=0}^n \frac{r}{^nC_r}$.Using the property $^nC_r = ^nC_{n-r}$,we can write:$b_n = \fr

Vedclass · Accepted Answer

$\frac{1}{2}na_n$

Vedclass · Answer

(C) Given $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$.Let $b_n = \sum_{r=0}^n \frac{r}{^nC_r}$.Using the property $^nC_r = ^nC_{n-r}$,we can write:$b_n = \frac{0}{^nC_0} + \frac{1}{^nC_1} + \frac{2}{^nC_2} + \dots + \frac{n}{^nC_n}$.Also,$b_n = \frac{n}{^nC_n} + \frac{n-1}{^nC_{n-1}} + \dots + \frac{0}{^nC_0} = \sum_{r=0}^n \frac{n-r}{^nC_r}$.Adding both expressions for $b_n$:$2b_n = \sum_{r=0}^n \frac{r + (n-r)}{^nC_r} = \sum_{r=0}^n \frac{n}{^nC_r} = n \sum_{r=0}^n \frac{1}{^nC_r}$.Since $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$,we have $2b_n = na_n$.Therefore,$b_n = \frac{1}{2}na_n$.

If $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$,then $\sum_{r=0}^n \frac{r}{^nC_r}$ equals

Similar Questions

All the letters of the word $TABLE$ are permuted and the strings of letters (may or may not have meaning) thus formed are arranged in dictionary order. Then the rank of the word $TABLE$ counted from the rank of the word $BLATE$ is

The value of $\sum\limits_{r = 1}^{15} {{r^2} \left( \frac{^{15}C_r}{^{15}C_{r - 1}} \right)}$ is equal to

There are $15$ stations on a train route and the train has to be stopped at exactly $5$ stations among these $15$ stations. If it stops at at least two consecutive stations,then the number of ways in which the train can be stopped is

If all the numbers which are greater than $6000$ and less than $10000$ are formed with the digits $3, 5, 6, 7, 8$ without repetition of the digits,then the difference between the number of odd numbers and the number of even numbers among them is

The largest power of $2$ that divides $\frac{200!}{100!}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module