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JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Sequences and Series

If

\frac{1}{2 \cdot 3 ^{10}} + \frac{1}{2 ^{2} \cdot 3 ^{9}} + \dots \frac{1}{2 ^{10} \cdot 3} = \frac{K}{2 ^{10} \cdot 3 ^{10}}

, then the remainder when K is divided by 6 is

(A)1
(B)2
(C)3
(D)5✓ Correct

Answer: (D)

Solution

The sum is

\sum_{k = 1}^{10} \frac{1}{2 ^{k} \cdot 3 ^{11 - k}} = \frac{1}{2 ^{10} \cdot 3 ^{10}} \sum_{k = 1}^{10} 2^{10 - k} \cdot 3^{k - 1}

. Thus

K = \sum_{k = 1}^{10} 2^{10 - k} \cdot 3^{k - 1} = 2^{9} + 2^{8} \cdot 3 + \dots + 3^{9}

, which is a geometric series with first term

2^{9}

and ratio

3/2

:

K = 2^{9} \cdot \frac{( 3/2 ) ^{10} - 1}{3/2 - 1} = 3^{10} - 2^{10}

. Now

3^{10} - 2^{10} = (3^{5})^{2} - (2^{5})^{2} = (243 - 32) (243 + 32) = 211 \times 275

. Since

211 = 35 \times 6 + 1

and

275 = 45 \times 6 + 5

, we have

K \equiv 1 \times 5 = 5 (mod 6)

.

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option

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