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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