Question

Let the general term of a series be $ (2k-1)\,\,(2k) $ $ (2k+1),\,\,k=1,2,3...,n. $ If the sum of first. $n$ terms is $ 24090, $ then $n$ is equal to

• A. $ 8 $
• B. $ 9 $
• C. $ 10 $
• D. $ 11 $

Answer 1

Answer: (C)

Given, general term of a series $ {{T}_{k}}=(2k-1)\,(2k)\,(2k+1). $
$ \Rightarrow $ $ {{T}_{k}}=\{{{(2k)}^{2}}-{{(1)}^{2}}\}\,\,(2k) $
$ \Rightarrow $ $ {{T}_{k}}=(4{{k}^{2}}-1)\,\,(2k) $
$ \Rightarrow $ $ {{T}_{k}}=8{{k}^{3}}-2k $
$ \Rightarrow $ $ {{T}_{n}}=8{{n}^{3}}-2n $ Then, $ {{S}_{n}}=\Sigma {{T}_{n}} $
$=\Sigma (8{{n}^{3}}-2\Sigma n) $
$ \Rightarrow $ $ {{S}_{n}}=8\Sigma {{n}^{3}}-2\Sigma n $
$=8{{\left( \frac{n(n+1)}{2} \right)}^{2}}-2\frac{n(n+1)}{2} $
$=8\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}-n(n+1) $ $ 2{{n}^{2}}{{(n+1)}^{2}}-n(n+1) $ But given, $ {{S}_{n}}=24090 $
$ \therefore $ $ 2{{n}^{2}}\,{{(n+1)}^{2}}-n(n+1)=24090 $
$ \Rightarrow $ $ 2{{n}^{2}}({{n}^{2}}+2n+1)-n(n+1)=24090 $
$ \Rightarrow $ $ 2({{n}^{4}}+2{{n}^{3}}+{{n}^{2}})-{{n}^{2}}-n=24090 $
$ \Rightarrow $ $ 2{{n}^{4}}+4{{n}^{3}}+2{{n}^{2}}-{{n}^{2}}-n=24090 $
$ \Rightarrow $ $ 2{{n}^{4}}+4{{n}^{3}}+{{n}^{2}}-n=24090 $
$ \Rightarrow $ $ n(2{{n}^{3}}+4{{n}^{2}}+n-1)=24090 $
Now, $ n=10 $ satisfy this equation. So, $ n=10 $