# Thread: Method of Differences

1. ## Method of Differences

Find the sum to $n$ terms of the series
$1(1!)+2(2!)+...+r(r!)+...$

Solution

Let,
$u_r=r(r!)$

If,
$f(r)=r!$

then,
$f(r+1)-f(r)$
$=(r+1)!-r!$
$=(r+1){\color{red}\text{r}}!-r!\quad\leftarrow$ Where is the ${\color{red}\text{r}}$ from?
$=r![(r+1)-1]$
$=r!(r)$
$=u_r$

Using the method of differences,
$1(1!)+2(2!)+...+n(n!)$
$=\sum \limits_{r=1}^nu_r$
$=\sum \limits_{r=1}^n[f(r+1)-f(r)]$
$=f(n+1)-f(1)$
$=(n+1)!-1$

2. Originally Posted by SengNee
Find the sum to $n$ terms of the series
$1(1!)+2(2!)+...+r(r!)+...$

Solution

Let,
$u_r=r(r!)$

If,
$f(r)=r!$

then,
$f(r+1)-f(r)$
$=(r+1)!-r!$
$=(r+1){\color{red}\text{r}}!-r!\quad\leftarrow$ Where is the ${\color{red}\text{r}}$ from?

Mr F answers: Do you know what (r+1)! means? It means (r+1)(r)(r-1)(r-2) ....... (2)(1). But (r)(r-1)(r-2) ...... (2)(1) is r! Therefore ......

$=r![(r+1)-1]$
$=r!(r)$
$=u_r$

Using the method of differences,
$1(1!)+2(2!)+...+n(n!)$
$=\sum \limits_{r=1}^nu_r$
$=\sum \limits_{r=1}^n[f(r+1)-f(r)]$
$=f(n+1)-f(1)$
$=(n+1)!-1$
..