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let `S_(n)` denote the sum of the cubes of the first n natural numbers and `s_(n)` denote the sum of the first n natural numbers , then `sum_(r=1)^(n)(S_(r))/(s_(r))` equals toA. `(n(n+1)(n+2))/(6)`B. `(n(n+1))/(2)`C. `(n^(2)+3n+2)/(2)`D. None of these |
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Answer» Correct Answer - A `sum _(r=1)^(n) (S_(r))/(S_(r))=(S_(1))/(S_(1))+(S_(2))/(S_(2))+(S_(3))/(S_(3))+...+(S_(n))/(S_(n))` Let `T_(n)` be the nth term the above series . `T_(n)=(S_(n))/(S_(n))=([(n(n+1))/(2)]^(2))/((n(n+1))/(2))` `=(n(n+1))/(2)=(1)/(2)[n^(2)=n]` `therefore `Sum of the above series `=sum T_(n)=(1)/(2)[sumn^(2)+sumn]` `=(1)/(2)[(n(n+1)(2n+1))/(6)+(n(n+1))/(2)]=(1)/(2).(n(n+1))/(2)[((2n+1))/(3)+1]` `=(1)/(4)m(n+1)[(2n+1+3)/(3)]=(1)/(4xx3)n(n+1)(2n+4)` `=(1)/(12)n(n+1)(2n+4)=(1)/(6)n(n+1)(n+2)` |
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