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JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Sequences and Series

If

\sum_{r = 1}^{13} {\frac{1}{s i n ( \frac{π}{4} + ( r - 1 ) \frac{π}{6} ) s i n ( \frac{π}{4} + \frac{r π}{6} )}} = a 3 + b

,

a, b \in Z

, then

a^{2} + b^{2}

is equal to :

(A)10
(B)2
(C)8✓ Correct
(D)4

Answer: (C)

Solution

For

A = \frac{π}{4} + (r - 1) \frac{π}{6}

and

B = \frac{π}{4} + r \frac{π}{6}

, note

B - A = \frac{π}{6}

, and

cot A - cot B = \frac{sin ( B - A )}{sin A sin B}

so

\frac{1}{sin A sin B} = \frac{cot A - cot B}{sin ( π /6 )} = 2 (cot A - cot B)

The sum telescopes:

r = 1 \sum 13 \frac{1}{sin A sin B} = 2 (cot \frac{π}{4} - cot (\frac{π}{4} + \frac{13 π}{6}))

Since

\frac{π}{4} + \frac{13 π}{6} = \frac{29 π}{12}

, and

cot

has period

π

, this equals

cot \frac{5 π}{12} = cot 7 5^{\circ} = 2 - 3

.

2 (1 - (2 - 3)) = 2 (3 - 1) = 23 - 2

So

a = 2

,

b = - 2

, and

a^{2} + b^{2} = 4 + 4 = 8

.

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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