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Math Help - Method of Differences

  1. #1
    Member SengNee's Avatar
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    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
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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 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
    ..
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