Question : Find the sum of $\left (1-\frac{1}{n+1} \right) + \left (1-\frac{2}{n+1} \right) + \left (1- \frac{3}{n+1} \right)+.....\left ( 1- \frac{n}{n+1} \right)$.
Option 1: $n$
Option 2: $\frac{n}{2}$
Option 3: $(n+1)$
Option 4: $\frac{(n+1)}{2}$
Correct Answer: $\frac{n}{2}$
Solution :
$\left (1-\frac{1}{n+1} \right) + \left (1-\frac{2}{n+1} \right) + \left (1- \frac{3}{n+1} \right)+.....\left (1- \frac{n}{n+1} \right)$
= $\frac{n}{n+1}+\frac{n-1}{n+1}+\frac{n-2}{n+1}+...........+\frac{1}{n+1}$
$= \frac{1}{(n+1)}\frac{[n(n+1)]}{2}$
$=\frac{n}{2}$
Hence, the correct answer is $\frac{n}{2}$.
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