Finding the sum of \(S_n=1\cdot1!+2\cdot2!+3\cdot3!+\cdots+n\cdot n!\), where \(n!=1\cdot2\cdot3\cdots n\).

Finding the sum of the following problem:

\(S_n=1\cdot1!+2\cdot2!+3\cdot3!+\cdots+n\cdot n!\)

where \(n!=1\cdot2\cdot3\cdots n\).

Solution:

In order to solve this kind of problem, we need to start with the general term of the sequence.

The general term of our problem is:

\(k\cdot k!\)

We can rewrite it as:

\(k\cdot k!=(k+1-1)k!=(k+1)k!-k!=(k+1)!-k!\)

Then we replace the value of \(k\) following our main problem:

\(1\cdot1!=2!-1!\)
\(2\cdot2!=3!-2!\)
\(3\cdot3!=4!-3!\)
\(\vdots\)
\(n\cdot n!=(n+1)!-n!\)

Adding all equations side by side:

\(S_n=(2!-1!)+(3!-2!)+(4!-3!)+\cdots+((n+1)!-n!)\)

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

Hence:

\(\boxed{S_n=(n+1)!-1!}\)

Solution by: Thin Sokkean