A) \[\frac{1}{2}a{{(a-1)}^{2}}\]
B) \[\frac{1}{2}a(a-1)(4a-1)\]
C) \[\frac{1}{2}(a-1)(2a-1)(4a-1)\]
D) None of these
Correct Answer: B
Solution :
[b] \[\begin{align} & AL_{1}^{2}+{{L}_{1}}M_{1}^{2}=({{a}^{2}}+{{1}^{2}})+\{{{(a-1)}^{2}}+{{1}^{2}}\} \\ & AL_{2}^{2}+{{L}_{2}}M_{2}^{2}=({{a}^{2}}+{{2}^{2}})+\{{{(a-2)}^{2}}+{{2}^{2}}\} \\ & ........................................................................ \\ & AL_{a-1}^{2}+{{L}_{a-1}}M_{a-1}^{2}={{a}^{2}}+{{(a-1)}^{2}}+\{{{1}^{2}}+{{(a-1)}^{2}}\} \\ \end{align}\] |
\[\therefore \] The required sum |
\[=(a-1){{a}^{2}}+\{{{1}^{2}}+{{2}^{2}}+...+{{(a-1)}^{2}}\}\] |
\[+2\{{{1}^{2}}+{{2}^{2}}+....+{{(a-1)}^{2}}\}\] |
\[=(a-1){{a}^{2}}+3.\,\,\frac{(a-1)a(2a-1)}{6}\] |
\[=a(a-1)\left\{ a+\frac{2a-1}{2} \right\}=\frac{a(a-1)(4a-1)}{2}\] |
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