Question

Find the sum of $\left( \sf{1 - \dfrac{1}{n + 1}} \right) + \left( \sf{1 - \dfrac{2}{n + 1}} \right) + \left( \sf{1 - \dfrac{3}{n + 1}} \right) + ... \left( \sf{1 - \dfrac{n}{n + 1}}$

$\bf{Options\; :}$

(a) n
(b) $\sf{\dfrac{1}{2} n}$
(c) (n + 1)
(d) $\sf{\dfrac{1}{2}(n + 1)}$
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Answer 1

Answer:

Step-by-step explanation:

To Find :-

Sum of :-

$\sf\left(1 - \dfrac{1}{n+1}\right)+\left(1-\dfrac{2}{n+1}\right)+\left(1-\dfrac{3}{n+1}\right)+..\left(1-\dfrac{n}{n+1}\right)$

Solution :-

$\sf\left(1 - \dfrac{1}{n+1}\right)+\left(1-\dfrac{2}{n+1}\right)+\left(1-\dfrac{3}{n+1}\right)+..\left(1-\dfrac{n}{n+1}\right)$

$\sf= 1-\dfrac{1}{n+1}+1-\dfrac{2}{n+1}+1-\dfrac{3}{n+1}+..1-\dfrac{n}{n+1}$

Putting all integers together and all fractions together:-

$\sf=1+1+1..+1-\dfrac{1}{n+1}-\dfrac{2}{n+1}-\dfrac{3}{n+1}..-\dfrac{n}{n+1}$

$\sf= \left(1\times n\right)-\left(\dfrac{1}{n+1}+\dfrac{2}{n+1}+\dfrac{3}{n+1}+..\dfrac{n}{n+1}\right)$

Since , Sum of 'n' one's = $\sf n\times 1$

$=n-\left(\dfrac{1+2+3+..n}{n+1}\right)$

$=n-\dfrac{n(n+1)}{2(n+1)}$

[Since, Sum of first 'n' terms = n(n+1)/2]

$= n - \dfrac{n}{2}$

$=\dfrac{2n-n}{2}$

$= \dfrac{n}{2}$

$=\dfrac{1}{2}n$

Since, option 'b' 1/2 n.

Answer 2

Correct answer is option ( b ) . For explanation , Kindly see the above attachment .