InterviewSolution
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InterviewSolution
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Find the sum of the series `(1^(2)+1)1!+(2^(2)+1)2!+(3^(2)+1)3!+ . .+(n^(2)+1)n!`. |
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Answer» Let `S_(n)=(1^(2)+1)1!+(2^(2)+1)2!+(3^(2)+1)3!+ . .+(n^(2)+1)n!` `therefore` nth term `T_(n)=(n^(2)+1)n!` `={(n+1)(n+2)-3(n+1)+2}n!` `T_(n)=(n+2)!-3(n+1)!+2n!` Putting `n=1,2,3,4, . .,n` then `T_(1)=3!-3*2!+2*1!` `T_(2)=4!-3*3!+2*2!` `T_(3)+%!-3*4!+2*3!` `T_(4)=6!-3*5!+2*4!` . . . . . . . . . . . . . . `T_(n-1)=(n+1)!-3n!+2(n-1)!` `T_(n)=(n+2)!-3(n+1)!+2n!` `therefore S_(n)=T_(1)+T_(2)+T_(3)+ . .. +T_(n)` `=(n+2)!-2(n+1)!` [the rest cancel out] `=(n+2)(n+1)!-2(n+1)!` `=(n+1)!(N=2-2)` `=n(n+1)!` |
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