A) \[8\]
B) \[9\]
C) \[10\]
D) \[11\]
Correct Answer: C
Solution :
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, \[n=10\]
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