Question

Find a closed form for

$S_n = 1 \cdot 1! + 2 \cdot 2! + \ldots + n \cdot n!.$
for integer $n \geq 1.$ Your response should have a factorial.

Answer 1

$S_n=1\cdot1!+2\cdot2!+\cdots+n\cdot n!$
$S_n=(1+1-1)\cdot1!+(2+1-1)\cdot2!+\cdots+(n+1-1)\cdot n!$
$S_n=(2\cdot1!-1\cdot1!)+(3\cdot2!-1\cdot2!)+\cdots+\bigg((n+1)n!-\cdot n!\bigg)$
$S_n=(2!-1!)+(3!-2!)+\cdots+\bigg((n+1)!-1\cdot n!\bigg)$
$S_n=-1!+(n+1)!$
$S_n=(n+1)!-1$

Answer 2

The closed form of the equation is: $S_n = (n + 1)! -1$

The equation is given as:

$S_n = 1 \cdot 1! + 2 \cdot 2! + \ldots + n \cdot n!.$

Rewrite the equation as:

$S_n = (1) \cdot 1! + (2) \cdot 2! + \ldots + (n) \cdot n!.$

Add 0 to the expressions in bracket

$S_n = (1 + 0) \cdot 1! + (2 + 0) \cdot 2! + \ldots + (n + 0) \cdot n!.$

Express 0 as 1 - 1

$S_n = (1 + 1 - 1) \cdot 1! + (2 + 1 - 1) \cdot 2! + \ldots + (n + 1 - 1) \cdot n!.$

Evaluate the sums

$S_n = (2 - 1) \cdot 1! + (3 - 1) \cdot 2! + \ldots + (n + 1 - 1) \cdot n!.$

Open the brackets

$S_n = 2 \cdot 1!- 1 \cdot 1! + 3 \cdot 2!- 1 \cdot 2! + \ldots + (n + 1 ) \cdot n!- 1 \cdot n!.$

Simplify

$S_n = (2!- 1!) + (3!- 2!) + \ldots +[ (n + 1 )!- 1 \cdot n!. ]$

Rewrite the above equation in terms of n

$S_n = (n + 1)! -1!$

Evaluate 1!

$S_n = (n + 1)! -1$

Hence, the closed form of the equation is:

$S_n = (n + 1)! -1$

Read more about factorials at:

https://brainly.com/question/8119212