# The value of $\left[\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\right)+\ldots \ldots+\left(1-\frac{n}{n+1}\right)\right]$ is

Given:

$\left[\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\right)+\ldots \ldots+\left(1-\frac{n}{n+1}\right)\right]$

To do:

We have to find the value of $\left[\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\right)+\ldots \ldots+\left(1-\frac{n}{n+1}\right)\right]$.

Solution:

$\begin{array}{l} \left[\left( 1-\frac{1}{n+1}\right) +\left( 1-\frac{2}{n+1}\right) +........+\left( 1-\frac{n}{n+1}\right)\right] =[ 1+1+....+1( n\ times)] -\frac{1}{n+1}[ 1+2+.....+n]\\ =n-\frac{1}{n+1}\left[\frac{n( n+1)}{2}\right]\\ =n-\frac{n}{2}\\ =\frac{2n-n}{2}\\ =\frac{n}{2} \end{array}$

Therefore, $\left[\left(1-\frac{1}{n+1}\right)+\left(1-\frac{2}{n+1}\right)+\ldots \ldots+\left(1-\frac{n}{n+1}\right)\right]=\frac{n}{2}$.

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Updated on: 10-Oct-2022

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