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JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Differentiation

Let

f (x) = \sum_{k = 1}^{10} kx^{k}, x \in R

. If

2 f (2) + f^{'} (2) = 119 (2)^{n} + 1

then n is equal to ______.

Answer: 10

Solution

Here

f (x) = \sum_{k = 1}^{10} k x^{k}

, so

f^{'} (x) = \sum_{k = 1}^{10} k^{2} x^{k - 1}

. Both are arithmetic\textendash geometric series; evaluating at

x = 2

gives

f (2) = \sum_{k = 1}^{10} k 2^{k} = 18434

and

f^{'} (2) = \sum_{k = 1}^{10} k^{2} 2^{k - 1} = 84989

. Hence

2 f (2) + f^{'} (2) = 36868 + 84989 = 121857

. Since

121857 = 119 \times 1024 + 1 = 119 \cdot 2^{10} + 1

, comparison with

119 \cdot 2^{n} + 1

gives

n = 10

.

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Integer answer

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