Question

PQRS is a square of length x, a natural number >1. Let $L_1,L_2,L_3,L_4\dots$be points on QR such that Q$L_1=L_1{L}_2=L_2{L}_3=L_3{L}_4\dots .=1$ and $M_1,M_2,M_3$ be points on RS such that R\(M_1=M_1M_2=M_2M_3. _{. . . . . .}=1\). Then, ${}_{n-1}^{a-1}\sum (P{L}_n^2+L_n{M}_n^2)$ is equal to?

• A. ½ x a x (a-1)²
• B. ½ x a x(a-1)x(4a-1)
• C. ½ x(a-1)x(2a-1)x(4a-1)
• D. ½ x(a+1)x(2a-1)x(4a-1)

Answer 1

The correct option is B ½ x a x(a-1)x(4a-1)

Option (b)

Assume a square of side 2 as follows with L${}_1$ as the midpoint of side QR and M${}_1$ as the mid-point of side RS

At n=1, the expression yields a value 5+2=7

Look in the answer options for 7, when n=1. Eliminate those answer options where this is not obtained

$\frac{1}{2}$*a*(a-1)${}^2$= 1

$\frac{1}{2}$ *a*(a-1)*(4a-1)= 7

$\frac{1}{2}$ *(a-1)*(2a-1)*(4a-1)= $\frac{21}{2}$

$\frac{1}{2}$ *(a+1)*(2a-1)*(4a-1)= $\frac{63}{2}$

a${}^2$*2n${}^2$ = 8