question_answer

ABCD is a square of lengths \[a,\,a\in N,a>1\]. Let \[{{L}_{1}},{{L}_{2}},{{L}_{3}},...\] be points BC such that \[B{{L}_{1}}={{L}_{1}}{{L}_{2}}={{L}_{2}}{{L}_{3}}=...=1\] and \[{{M}_{1}},{{M}_{2}},{{M}_{3}},...\] be points on CD such that \[C{{M}_{1}}={{M}_{1}}{{M}_{2}}={{M}_{2}}{{M}_{3}}=...=1\]. Then, \[\sum\limits_{n=1}^{a-1}{(AL_{n}^{2}+{{L}_{n}}M_{n}^{2})}\] is equal to

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

\[=(a-1){{a}^{2}}+\{{{1}^{2}}+{{2}^{2}}+...+{{(a-1)}^{2}}\}\]

\[+2\{{{1}^{2}}+{{2}^{2}}+....+{{(a-1)}^{2}}\}\]

\[=(a-1){{a}^{2}}+3.\,\,\frac{(a-1)a(2a-1)}{6}\]

\[=a(a-1)\left\{ a+\frac{2a-1}{2} \right\}=\frac{a(a-1)(4a-1)}{2}\]