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# Engineering

The sum

\frac{3\times 1^{3}}{1^{2}}+\frac{5\times (1^{3}+2^{3})}{1^{2}+2^{2}}+\frac{7\times (1^{3}+2^{3}+3^{3})}{1^{2}+2^{2}+3^{2}}+.......... 

upto 10th term , is :

  • Option 1)

    680

  • Option 2)

    600

  • Option 3)

    660

  • Option 4)

    620

Answers (1)
V Vakul

Given,

S_n=\frac{3\times 1^{3}}{1^{2}}+\frac{5\times (1^{3}+2^{3})}{1^{2}+2^{2}}+\frac{7\times (1^{3}+2^{3}+3^{3})}{1^{2}+2^{2}+3^{2}}+..........

general term will be 

T_n=\frac{(3+(n-1)2)(1^{3}+2^{3}+3^{3}+...........)}{(1^{2}+2^{2}+3^{2}+........)}

      =\frac{(2n+1)[\frac{n(n+1)}{2}]^{2}}{\frac{n(n+1)(2n+1)}{6}}=\frac{3}{2}n(n+1)

S_n=\sum T_n=\frac{3}{2}\sum_{n=1}^{10}(n^{2}+n)

                          =\frac{3}{2}[\frac{10\times 11\times 21}{6}+\frac{10\times 11}{2}]

                          =660

So, correct option is (3).


Option 1)

680

Option 2)

600

Option 3)

660

Option 4)

620

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