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Summation by Sigma Method

ABCD is a squ...

Question

ABCD is a square of length a, $a ϵ N$ , a>1. Let $L_{1}$ , $L_{2}$ , $L_{3}$ ,... be points 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 $C M_{1} = M_{1} M_{2} = M_{2} M_{3} = . . . = 1$ . Then $\sum_{n = 1}^{a - 1} (A {L_{n}}^{2} + L_{n} {M_{n}}^{2})$ is equal to

A

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

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B

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

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C

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

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D

none of these

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Solution

The correct option is B $\frac{1}{2} a (a - 1) (4 a - 1)$
$A {L_{n}}^{2} = a^{2} + n^{2} {L_{n} M_{n}}^{2} = {(a - n)}^{2} + n^{2} A {L_{n}}^{2} + {L_{n} M_{n}}^{2} = 2 a^{2} + {3 n}^{2} - 2 n a$
Thus, the summation is:
$2 a^{2} (a - 1) + 3. \frac{(a - 1) (a) (2 a - 1)}{6} - 2 a . \frac{(a - 1) (a)}{2} = \frac{a}{2} . (a - 1) . (4 a + 2 a - 1 - 2 a) = \frac{1}{2} a . (a - 1) . (4 a - 1)$

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Similar questions

A B C D

is a square of length

a, a \in N, a > 1.

L_{1}, L_{2}, L_{3}, \dots

B C

B L_{1} = L_{1} L_{2} = L_{2} L_{3} = \dots = 1

M_{1}, M_{2}, M_{3}, \dots

C D

C M_{1} = M_{1} M_{2} = M_{2} M_{3} = \dots = 1.

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

A B C D

is a square of length

a, a \in N, a > 1

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

be point on BC such that

B L_{1} = L_{1} L_{2} = L_{2} L_{3} = . . . . . = 1

M_{1}, M_{2}, M_{3} . . . . . . . . .

be points on CD such that

C M_{1} = M_{1} M_{2} = . . . . . . . . . . . = 1

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

Q. PQRS is a square of length a, a natural number >1. Let

L_{1}, L_{2}, L_{3}, L_{4} \dots

be points on QR such that Q

L_{1} = L_{1} L_{2} = L_{2} L_{3} = L_{3} L_{4} \dots . = 1

M_{1}, M_{2}, M_{3}

be points on RS such that R

M_{1} = M_{1} M_{2} = M_{2} M_{3} . . . .

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

l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2}; l_{3}, m_{3}, n_{3}

are real quantities satisfying the six relations

l_{1}^{2} + m_{1}^{2} + n_{1}^{2} = l_{2}^{2} + m_{2}^{2} + n_{2}^{2} = l_{3}^{2} + m_{3}^{2} + n_{3}^{2} = 1

l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} = l_{3} l_{1} + m_{3} m_{1} + n_{3} n_{1} = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0

∣ ∣ ∣ ∣ \begin{matrix} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} l_{3} & m_{3} & n_{3} \end{matrix} ∣ ∣ ∣ ∣ = \pm k

. Find the value of

k

A B C

is a right angled

△

∠ B = 90^{\circ}

B C = a

n

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

A B

A B

is divided into

(n + 1)

equal parts and

L_{1} M_{1}, L_{2} M_{2}, . . . L_{n} M_{n}

are line segments parallel to

B C

M_{1}, M_{2}, M_{3}, . . ., M_{n}

A C

, then the sum of the lengths of

L_{1} M_{1}, L_{2} M_{2}, . . . L_{n} M_{n}

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