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Find the sum of n terms of the series whose nth terms is (i) `n(n-1)(n+1)`. |
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Answer» (i) We have, `T_(n)=n(n-1)(n+1)=n^(3)-n` ` therefore " sum of n terms " S_(n)= sumT_(n)` `2=sumn^(3)+2sumn^(2)+sumn` `={(n(n+1))/(2)}^(2)-{(n(n+1))/(2)}` `=(n(n+1))/(2)-{(n(n+1))/(2)-1}` `=(n(n+1)(2)(n-1)(n+2))/(4)` |
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