ABCD is a square of length a- a - N- a 1- Let L1- L2- L3- - be points on BC such that BL1 - L1 L2 - L2 L3 - - - 1 and M1 - M2- M3- - are points on CD such that CM1 - M1M2 - M2 M3 - - - 1- Then - n - 1 a - 1ALn 2 -LnMn 2 is equal to

(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 nbsp; 1)^2 = [a^2 +(a 1 )^ ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

ABCD is a squ...

Question

$A B C D$ is a square of length $a, a \in N, a > 1.$ Let $L_{1}, L_{2}, L_{3}, \dots$ be points on $B C$ such that $B L_{1} = L_{1} L_{2} = L_{2} L_{3} = \dots = 1$ and $M_{1}, M_{2}, M_{3}, \dots$ are points on $C D$ such that $C M_{1} = M_{1} M_{2} = M_{2} M_{3} = \dots = 1.$ Then $a - 1 \sum n = 1 (A {L_{n}}^{2} + L_{n} {M_{n}}^{2})$ is equal to

\frac{1}{2} a (a - 1)

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\frac{1}{2} (a - 1) (2 a - 1) (4 a - 1)

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\frac{1}{2} a (a - 1) (4 a - 1)

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\frac{1}{2} (a - 1)^{2} (2 a - 1)

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Solution

The correct option is C $\frac{1}{2} a (a - 1) (4 a - 1)$

$(A L_{1})^{2} + (L_{1} M_{1})^{2} = [a^{2} + 1^{2}] + [(a - 1)^{2} + 1^{2}] (A L_{2})^{2} + (L_{2} M_{2})^{2} = [a^{2} + 2^{2}] + [(a - 2)^{2} + 2^{2}] (A L_{a - 1})^{2} + (L_{a - 1} M_{a - 1})^{2} = [a^{2} + (a - 1)^{2}] + [(1)^{2} + (a - 1)^{2}]$

Therefore the required sum will be,
$= (a - 1) a^{2} + [1^{2} + 2^{2} + 3^{2} + \dots + (a - 1)^{2}] + 2 [1^{2} + 2^{2} + 3^{2} + \dots + (a - 1)^{2}] = (a - 1) a^{2} + 3 [1^{2} + 2^{2} + 3^{2} + \dots + (a - 1)^{2}] = (a - 1) a^{2} + 3 \times \frac{(a - 1) (a) (2 a - 1)}{6} = \frac{1}{2} a (a - 1) (4 a - 1)$

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Similar questions

A B C D

is a square of length

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and

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ABCD is a square of length a,a∈N,a>1. Let L1,L2,L3,⋯ be points on BC such that BL1=L1L2=L2L3=⋯=1 and M1,M2,M3,⋯ are points on CD such that CM1=M1M2=M2M3=⋯=1. Then a−1∑n=1(ALn2+LnMn2) is equal to