Question

please solve this i will mark the brainliyest but give the solution with explanation otherwise I will report it​

Answer 1

Answer :

The series is in AP.

Where ,

First term = a = $(1-\frac{1}{n} )$

Common difference = d = $(1-\frac{2}{n} )-(1-\frac{1}{n} )=1-\frac{2}{n} -1+\frac{1}{n} =\frac{-1}{n}$

=> Sum of n terms in AP , $S_n$

$=>\frac{n}{2} [2a+(n-1)d]\\\\=>\frac{n}{2} [2(1-\frac{1}{n} )+(n-1)\frac{-1}{n} ]\\\\=>\frac{n}{2} [2-\frac{2}{n}-1+\frac{1}{n} ]\\\\=>\frac{n}{2} [1-\frac{1}{n} ]\\\\=>\frac{n}{2} [\frac{n-1}{n} ]\\\\=>\frac{n-1}{2}$

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