A) 10
B) 7
C) 8
D) 9
E) 6
Correct Answer: E
Solution :
Given series is \[{{2}^{3}}+{{4}^{3}}+{{6}^{3}}+...\] \[\therefore \] \[{{T}_{n}}={{(2n)}^{3}}=8{{n}^{3}}\] \[\therefore \] \[\Sigma {{T}_{n}}=8\Sigma {{n}^{3}}=\frac{8{{n}^{2}}{{(n+1)}^{2}}}{4}\] \[\Rightarrow \] \[3528=2{{n}^{2}}{{(n+1)}^{2}}\] (given) \[\Rightarrow \] \[n=6\]You need to login to perform this action.
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