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JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Binomial Theorem

2 \cdot^{20} C_{0} + 5 \cdot^{20} C_{1} + 8 \cdot^{20} C_{2} + 11 \cdot^{20} C_{3} + \dots \cdot + 62 \cdot^{20} C_{20}

is equal to

(A) $2^{23}$
(B) $2^{26}$
(C) $2^{24}$
(D) $2^{25}$ ✓ Correct

Answer: (D)

Solution

Write the general term as

(3 r + 2) (r 20)

and split the sum into

3 r = 0 \sum 20 r (r 20) + 2 r = 0 \sum 20 (r 20)

. Using

r (r 20) = 20 (r - 1 19)

, the first sum is

20 \cdot 2^{19}

and the second is

2^{20}

. So the total is

3 \cdot 20 \cdot 2^{19} + 2 \cdot 2^{20} = 60 \cdot 2^{19} + 2^{21} = 30 \cdot 2^{20} + 2 \cdot 2^{20} = 32 \cdot 2^{20} = 2^{25}

.

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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