The value of sumlimits_{r = 0}^{n - 1} {frac{ | vedclass

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

Question

(B) Let $S = \sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $.Using the property $\frac{^n{C_{r+1}}}{^n{C_r}} = \frac{n

Vedclass · Accepted Answer

$\frac{n}{2}$

Vedclass · Answer

(B) Let $S = \sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $.Using the property $\frac{^n{C_{r+1}}}{^n{C_r}} = \frac{n-r}{r+1}$,we can rewrite the term inside the summation:$\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}} = \frac{1}{1 + \frac{^n{C_{r+1}}}{^n{C_r}}} = \frac{1}{1 + \frac{n-r}{r+1}} = \frac{r+1}{n+1}$.Now,substitute this back into the summation:$S = \sum\limits_{r = 0}^{n - 1} \frac{r+1}{n+1} = \frac{1}{n+1} \sum\limits_{r = 0}^{n - 1} (r+1)$.Expanding the sum: $\sum\limits_{r = 0}^{n - 1} (r+1) = 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$.Therefore,$S = \frac{1}{n+1} \cdot \frac{n(n+1)}{2} = \frac{n}{2}$.

The value of $\sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $ equals

Similar Questions

Eleven books consisting of $5$ Mathematics,$4$ Physics,and $2$ Chemistry are placed on a shelf. The number of possible ways of arranging them on the assumption that the books of the same subject are all together is

The number of ways in which first,second and third prizes can be given to $5$ competitors is

The total number of ways of selecting six coins out of $20$ one-rupee coins,$10$ fifty-paise coins,and $7$ twenty-five-paise coins is:

The number of ways in which an examiner can assign $30$ marks to $8$ questions,giving not less than $2$ marks to any question,is

In an examination,there are three multiple-choice questions,and each question has $4$ choices. The number of ways in which a student can fail to get all answers correct is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module