Question

Find the sum of first n terms of the series

1 + 2. 2! + 3. 3! + 4. 4! + --------​

Answer 1

$\large\underline{\sf{Solution-}}$

Given series is

$\rm \: 1 + 2.2! + 3.3! + 4.4! + - - - n \: terms$

can be rewritten as

$\rm \: = \: 1.1! + 2.2! + 3.3! + 4.4! + - - - n \: terms$

So, nth term of the series is given by

$\rm \: t_{n} = n. \: n!$

can be further rewritten as

$\rm \: t_{n} = (n + 1 - 1). \: n!$

$\rm \: t_{n} = (n + 1).n! \: - \: n!$

On substituting n = 1, 2, 3, ------, n, we get

$\rm \: t_{1} = (1 + 1).1! \: - \: 1! = 2.1! - 1! = 2! - 1!$

$\rm \: t_{2} = (2 + 1).2! \: - \: 2! = 3.2! - 2! = 3! - 2!$

$\rm \: t_{3} = (3 + 1).3! \: - \: 3! = 4.3! - 3! = 4! - 3!$

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$\rm \: t_{n} = (n + 1).n! \: - \: n!$

Now, we know Sum of n terms is

$\rm \: S_n \: = \: \displaystyle\sum_{n=1}^n\rm t_{n}$

$\rm \: = \: t_{1} + t_{2} + t_{3} + - - - - + t_{n}$

$\rm \: = \: (2! - 1!) + (3! - 2!) + (4! - 3!) + - - - + (n + 1)! - n!$

$\rm \: = \: (n + 1)! \: - \: 1!$

$\rm \: = \: (n + 1)! - 1$

Hence,

$\boxed{\tt{ \rm \: 1 + 2.2! + 3.3! + 4.4! + - - - n \: terms = (n + 1)! - 1\: }}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ \displaystyle\sum_{n=1}^n\rm n = 1 + 2 + 3 + - - + n \: = \frac{n(n + 1)}{2} \: }}$

$\boxed{\tt{ \displaystyle\sum_{n=1}^n\rm {n}^{2} = {1}^{2} + {2}^{2} + {3}^{2} + - - + {n}^{2} \: = \frac{n(n + 1)(2n + 1)}{6} \: }}$

$\boxed{\tt{ \displaystyle\sum_{n=1}^n\rm {n}^{3} = {1}^{3} + {2}^{3} + {3}^{3} + - - + {n}^{3} \: = {\bigg[\dfrac{n(n + 1}{2} \bigg]}^{2} \: }}$

Answer 2

[Refer to the above attachment]

Hope you have satisfied. ⚘