Sum of the series - r-1 n r 2 -1 r- is

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Sum of the se...

Question

Sum of the series $n \sum r = 1 (r^{2} + 1) r!$ is

(n + 1)!

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(n + 2)! - 1

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n \cdot (n + 1)!

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n \cdot (n + 2)!

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Solution

The correct option is C $n \cdot (n + 1)!$
We check it by putting the different values of n
Let n=2
$\sum_{r = 1}^{2} (r^{2} + 1) . r!$ = $2.1! + 5.2!$ = $12$

Now check it with the options given
$A)$ $(n + 1)!$ $=$ $3!$ $= 6$

$B)$ $(n + 2)! - 1$ $=$ $4! - 1$ $= 23$

$C)$ $n . (n + 1)!$ $=$ $2.3!$ $= 12$

$D)$ $n . (n + 2)!$ $=$ $2.4!$ $= 48$

Hence C) is the correct answer

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Sum of the series n∑r=1(r2+1)r! is

The correct option is C n⋅(n+1)!We check it by putting the different values of nLet n=2∑2r=1(r2+1).r!=2.1!+5.2!=12Now check it with the options givenA) (n+1)! = 3! =6B) (n+2)!−1 = 4!−1 =23C) n.(n+1)! = 2.3! =12D) n.(n+2)! = 2.4! =48Hence C) is the correct answer

Sum of the series $n \sum r = 1 (r^{2} + 1) r!$ is