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  4. If the sum of n terms of the series 23+43+ 63 + ... is 3528, then n equals to :

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Q. If the sum of $ n $ terms of the series $ {{2}^{3}}+{{4}^{3}}+ $ $ {{6}^{3}} $ + ... is 3528, then $ n $ equals to :

KEAMKEAM 2004

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 $