Question

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$\dashrightarrow \ \sf \dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} \ - \ \dfrac{1}{(n - 2)!}$


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Answer 1

Answer:

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$\dashrightarrow \ \sf \dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} \ - \ \dfrac{1}{(n - 2)!}$

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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)(1−n+11)+(1−n+12)+(1−n+13)+..(1−n+1n)

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)(1−n+11)+(1−n+12)+(1−n+13)+..(1−n+1n)

\sf= 1-\dfrac{1}{n+1}+1-\dfrac{2}{n+1}+1-\dfrac{3}{n+1}+..1-\dfrac{n}{n+1}=1−n+11+1−n+12+1−n+13+..1−n+1n

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}=1+1+1..+1−n+11−n+12−n+13..−n+1n

\sf= \left(1\times n\right)-\left(\dfrac{1}{n+1}+\dfrac{2}{n+1}+\dfrac{3}{n+1}+..\dfrac{n}{n+1}\right)=(1×n)−(n+11+n+12+n+13+..n+1n)

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

=n-\left(\dfrac{1+2+3+..n}{n+1}\right)=n−(n+11+2+3+..n)

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

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

= n - \dfrac{n}{2}=n−2n

=\dfrac{2n-n}{2}=22n−n

= \dfrac{n}{2}=2n

=\dfrac{1}{2}n=21n

Since, option 'b' 1/2 n.

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