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Thread: Method of Differences

  1. #1
    Member SengNee's Avatar
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    Method of Differences

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


    Solution

    Let,
    $\displaystyle u_r=r(r!)$

    If,
    $\displaystyle f(r)=r!$

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

    Using the method of differences,
    $\displaystyle 1(1!)+2(2!)+...+n(n!)$
    $\displaystyle =\sum \limits_{r=1}^nu_r$
    $\displaystyle =\sum \limits_{r=1}^n[f(r+1)-f(r)]$
    $\displaystyle =f(n+1)-f(1)$
    $\displaystyle =(n+1)!-1$
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by SengNee View Post
    Find the sum to $\displaystyle n$ terms of the series
    $\displaystyle 1(1!)+2(2!)+...+r(r!)+...$


    Solution

    Let,
    $\displaystyle u_r=r(r!)$

    If,
    $\displaystyle f(r)=r!$

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

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

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