The sum of the series $(1^{2} + 1) .1! + (2^{2} + 1) .2! + (3^{2} + 1) .3! + . . . . . . + (n^{2} + 1) . n!$ is

(n+1).(n+2)!

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n.(n+1)!

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(n+1).(n+1)!

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none of these

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Solution

The correct option is A (n+1).(n+2)!
We have
$S_{n} = (1^{2} + 1) .1! + (2^{2} + 1) .2! + (3^{2} + 1) .3! + . . . . . + (n^{2} + 1) . n!$

$= 1^{2} .1! +! + 2^{2} .2! + 2! + 3^{2} .3! + 3! + . . . . . . . + n^{2} n! + n!$

$= (n + 1) (n + 2)!$
Hence, the option $(A)$ is correct answer.

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Similar questions

Q. The sum of the series

(1^{2} + 1) .1! + (2^{2} + 1) .2! + (3^{2} + 1) .3! + . . . . + (n^{2} + 1) . n! i s :

1 + 2 + 3 + 4 + . . . . . + n = ?

1 + 2 + 3 + 4 + . . . . . + (n - 1) = ?

1 + 1 + 1 + 1 + . . . . . + n = ?

1 + 1 + 1 + 1 + . . . . . . + (n - 1) = ?

1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + n^{2} = ?

1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + {(n - 1)}^{2} = ?

Q. Show that

\frac{1 \times 2^{2} + 2 \times 3^{2} + . . . + n \times {(n + 1)}^{2}}{1^{2} \times 2 + 2^{2} \times 3 + . . . + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}

Show that

$\frac{1 \times 2^{2} + 2 \times 3^{2} + \dots \dots + n \times (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + \dots \dots + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}$

Q. The sum of series

\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + . . . .

upto n terms is

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The sum of the series (12+1).1!+(22+1).2!+(32+1).3!+......+(n2+1).n! is

The correct option is A (n+1).(n+2)!We have Sn=(12+1).1!+(22+1).2!+(32+1).3!+.....+(n2+1).n! =12.1!+!+22.2!+2!+32.3!+3!+.......+n2n!+n! =(n+1)(n+2)!Hence, the option (A) is correct answer.

The sum of the series $(1^{2} + 1) .1! + (2^{2} + 1) .2! + (3^{2} + 1) .3! + . . . . . . + (n^{2} + 1) . n!$ is

The correct option is A (n+1).(n+2)!
We have
$S_{n} = (1^{2} + 1) .1! + (2^{2} + 1) .2! + (3^{2} + 1) .3! + . . . . . + (n^{2} + 1) . n!$

$= 1^{2} .1! +! + 2^{2} .2! + 2! + 3^{2} .3! + 3! + . . . . . . . + n^{2} n! + n!$

$= (n + 1) (n + 2)!$
Hence, the option $(A)$ is correct answer.