Question

ABCD is a square of lengths a, a∈N, a>1. Let L1, L2, L3,... be points on BC such that \[B{L_1} = {L_1}{L_2} = {L_2}{L_3} = ... = 1\] and M1, M2, M3,... be points on CD such that \[C{M_1} = {M_1}{M_2} = {M_2}{M_3} = ... = 1\]. Then\[\sum\limits_{n = 1}^{a - 1} {(A{L_n}^2 + {L_n}{M_n}^2)} \]is equal to
A) \[\dfrac{1}{2}a{(a - 1)^2}\]
B) \[\dfrac{1}{2}a(a - 1)(4a - 1)\]
C) \[\dfrac{1}{2}a(a - 1)(2a - 1)(4a - 1)\]
D) None of these

Answer 1

Hint: In this question we have to find the value of \[\sum\limits_{n = 1}^{a - 1} {(A{L_n}^2 + {L_n}{M_n}^2)} \]. By using Pythagoras theorem first find \[A{L_n}^2\] and \[{L_n}{M_n}^2\]. Now apply summation function on \[(A{L_n}^2 + {L_n}{M_n}^2)\].

Formula used: Formula of summation is
\[\sum\limits_{n = 1}^\infty {({a^n})} = a + {a^2} + {a^3}...........\]

Complete step by step solution: ABCD is a square L1,L2,L3,... be points on BC such that \[B{L_1} = {L_1}{L_2} = {L_2}{L_3} = ... = 1\] and M1,M2,M3,... be points on CD such that \[C{M_1} = {M_1}{M_2} = {M_2}{M_3} = ... = 1\]
By using Pythagoras theorem we can calculate
\[{(A{L_n})^2} = {(AB)^2} + {(B{L_n})^2} = {a^2} + {n^2}\]
Similarly
\[{({L_n}{M_n})^2} = {(a - n)^2} + {n^2}\]
Now use property of summation
\[\sum\limits_{n = 1}^{a - 1} {(A{L_n}^2 + {L_n}{M_n}^2)} = \sum\limits_{n = 1}^{a - 1} {({a^2} + {n^2} + {{(a - n)}^2} + {n^2})} \]
\[\sum\limits_{n = 1}^{a - 1} {(A{L_n}^2 + {L_n}{M_n}^2)} = \sum\limits_{n = 1}^{a - 1} {({a^2} + {n^2} + {a^2} + {n^2} - 2an + {n^2})} \]
\[\sum\limits_{n = 1}^{a - 1} {(A{L_n}^2 + {L_n}{M_n}^2)} = \sum\limits_{n = 1}^{a - 1} {(2{a^2} + 3{n^2} - 2an)} \]
\[\sum\limits_{n = 1}^{a - 1} {(A{L_n}^2 + {L_n}{M_n}^2)} = \sum\limits_{n = 1}^{a - 1} {2{a^2}} + \sum\limits_{n = 1}^{a - 1} {3{n^2}} - \sum\limits_{n = 1}^{a - 1} {2an} \]
\[2{a^2}\sum\limits_{n = 1}^{a - 1} 1 + 3\sum\limits_{n = 1}^{a - 1} {{n^2}} - 2a\sum\limits_{n = 1}^{a - 1} n \]
\[2{a^2}(a - 1) + \dfrac{{3(a - 1)a(2a - 1)}}{6} - \dfrac{{2a(a - 1)a}}{2}\]
\[(a - 1)a\{ \dfrac{{(2a - 1)}}{2} - a + 2a\} \]
Now required value is
\[\sum\limits_{n = 1}^{a - 1} {(A{L_n}^2 + {L_n}{M_n}^2)} = \dfrac{1}{2}a(a - 1)(4a - 1)\]


Thus, Option (B) is correct.

Note: Square is a geometrical figure having four sides and all sides are equal in length. In square each angle is 90 degree. Each opposite sides is parallel to each other. There are total four vertexes in square. Area of square is equal to square of its side. Diagonals of square are perpendicular to each other. Diagonal of square divide square into equal triangle. Perimeter of square is equal to four times its sides.