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ABCD is a square of length a,a N,a>1 . Let L_{1},L_{2},L_{3}

Question

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

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

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

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

None of these

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Owen Discussed

A BC D

is a square of length

a, a \in N, a > 1

. Let

L_{1}, L_{2}, L_{3} ......,

be point on 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} = ........... = 1

. Then

n = 1 \sum a - 1 (A L_{2}^{2} + L_{n} M_{n}^{2})

is equal to

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Text solutionVerified

(A L_{n})^{2} = (A B)^{2} + (B L_{n})^{2} = a^{2} + n^{2} (L_{n} M_{n})^{2} = (a - n)^{2} + n^{2} n = 1 \sum a - 1 ((A L_{n})^{2} + (L_{n} M_{n})^{2}) = n = 1 \sum a - 1 (a^{2} + n^{2} + n^{2} + n^{2} - 2 an + a^{2}) n = 1 \sum a - 1 (3 n^{2} - 2 an + 2 a^{2}) = 3 n = 1 \sum a - 1 n^{2} - 2 a n = 1 \sum a - 1 n + 2 a^{2} n = 1 \sum a - 1 (1) \frac{3 ( a - 1 ) a ( 2 a - 1 )}{6} - \frac{2 a ( a - 1 ) a}{2} + 2 a^{2} (a - 1) (a - 1) a [\frac{( 2 a - 1 )}{2} - a + 2 a] = \frac{1}{2} a (a - 1) (4 a - 1)

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Question Text	$A BC D$ is a square of length $a, a \in N, a > 1$ . Let $L_{1}, L_{2}, L_{3} ......,$ be point on 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 $C M_{1} = M_{1} M_{2} = ........... = 1$ . Then $n = 1 \sum a - 1 (A L_{2}^{2} + L_{n} M_{n}^{2})$ is equal to
Answer Type	Text solution:1
Upvotes	150