A) \[{{n}^{3}}+{{n}^{2}}+1\]
B) \[{{n}^{3}}-{{n}^{2}}+1\]
C) \[{{n}^{3}}-{{n}^{2}}\]
D) \[{{n}^{3}}+{{n}^{2}}\]
Correct Answer: C
Solution :
Given,\[{{a}_{0}}=I\] and \[{{a}_{n+1}}=3{{n}^{2}}+n+{{a}_{n}}\] \[\Rightarrow \] \[{{a}_{n}}=3{{(n-1)}^{2}}+(n-1)+{{a}_{n-1}}\] Now, put\[n=1,\,\,2,\,\,3,...n\] We get, \[{{a}_{1}}=0+{{a}_{0}}=0+1=1\] \[\Rightarrow \] \[{{a}_{2}}=4+{{a}_{1}}\] \[\Rightarrow \] \[{{a}_{3}}=14+{{a}_{2}}\] \[\Rightarrow \] \[{{a}_{4}}=30+{{a}_{3}}\] ?????????.. ?????????.. ?????????.. \[\Rightarrow \] \[{{a}_{n-1}}=3{{(n-2)}^{2}}+(n-2)+{{a}_{n-2}}\] \[\Rightarrow \] \[{{a}_{n}}=3{{(n-1)}^{2}}+(n-1)+{{a}_{n-1}}\] On adding, we get \[{{a}_{n}}=(1+4+14+30+...+3{{(n-1)}^{2}}+(n-1))\] \[\Rightarrow \] \[{{a}_{n}}=\Sigma 3({{n}^{2}}+1-2n)+(n-1)\] \[\Rightarrow \] \[{{a}_{n}}=\Sigma (3{{n}^{2}}+3-6n+n-1)\] \[\Rightarrow \] \[{{a}_{n}}=\Sigma (3{{n}^{2}}-5n+2)\] \[\Rightarrow \] \[{{a}_{n}}=3\Sigma {{n}^{2}}-5{{\Sigma }_{n}}+2\Sigma 1\] \[\Rightarrow \] \[{{a}_{n}}=\frac{3n(n+1)(2n+1)}{6}-\frac{5n(n+1)}{2}+2n\] \[\Rightarrow \] \[{{a}_{n}}=\frac{n}{2}(n+1)\{2n+1-5)+2n\] \[\Rightarrow \] \[{{a}_{n}}=\frac{n}{2}(n+1)(2n-4)+2n\] \[=n(n+1)(n-2)+2n\] \[\Rightarrow \] \[{{a}_{n}}=({{n}^{2}}+n)(n-2)+2n\] \[={{n}^{3}}-2{{n}^{2}}+{{n}^{2}}-2n+2n\] \[\Rightarrow \] \[{{a}_{n}}={{n}^{3}}-{{n}^{2}}\] \[(\because \,\,n\ge \theta )\]
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