Find the sum of left(1-frac{1}{n+1}right)+lef | vedclass

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

Question

(B) The given expression is $\sum_{k=1}^{n} \left(1-\frac{k}{n+1}\right)$.This can be expanded as $\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\

Vedclass · Accepted Answer

$\frac{1}{2} n$

Vedclass · Answer

(B) The given expression is $\sum_{k=1}^{n} \left(1-\frac{k}{n+1}\right)$.This can be expanded as $\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\right)+\cdots+\left(1-\frac{n}{n+1}\right)$.There are $n$ terms in this series,so we can write it as $n - \left(\frac{1}{n+1} + \frac{2}{n+1} + \cdots + \frac{n}{n+1}\right)$.Factoring out the denominator,we get $n - \frac{1}{n+1} (1+2+3+\cdots+n)$.Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we substitute this into the expression:$n - \frac{1}{n+1} \cdot \frac{n(n+1)}{2}$.Canceling $(n+1)$ from the numerator and denominator,we get $n - \frac{n}{2}$.Thus,the sum is $\frac{n}{2}$.

Find the sum of $\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\right)+\left(1-\frac{3}{n+1}\right)+\cdots+\left(1-\frac{n}{n+1}\right)$

Similar Questions

If $A$ is an arithmetic mean between two numbers and $S$ is the sum of $n$ arithmetic means between the same two numbers,then:

Find the common difference of an $A.P.$ whose first term is $100$ and the sum of whose first six terms is five times the sum of the next six terms.

The sum of $1 + 3 + 5 + 7 + \dots$ up to $n$ terms is:

The $9^{th}$ term of the series $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$ will be

The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots$ up to the $11^{th}$ term is

Vedclass Test Series

Exam Paper Generator

Online Exam Module