Let f ( x ) = k = 1 10 k x k , x . If 2 f ( 2 ) + f ' ( 2 ) = 119 ( 2 ) n + 1 , then n is equal to _________ . [2023]

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Solution

Q.
Let $f (x) = \sum_{k = 1}^{10} k x^{k}, x \in ℝ .$ If $2 f (2) + f' (2) = 119 {(2)}^{n} + 1,$ then $n$ is equal to _________ . [2023]

Ans.
(10)

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

$f (x) = x + 2 x^{2} + 3 x^{3} + \dots + 10 x^{10}$

$x \cdot f (x) = x^{2} + 2 x^{3} + 3 x^{4} + \dots + 9 x^{10} + 10 x^{11}$

$f (x) (1 - x) = x + x^{2} + x^{3} + \dots + x^{10} - 10 x^{11}$

$f (x) = \frac{x (1 - x^{10})}{{(1 - x)}^{2}} - \frac{10 x^{11}}{1 - x}$

$f (x) = \frac{x - x^{11} - 10 x^{11} (1 - x)}{{(1 - x)}^{2}}$

$f (x) = \frac{x - x^{11} - 10 x^{11} + 10 x^{12}}{{(1 - x)}^{2}} = \frac{10 x^{12} - 11 x^{11} + x}{{(1 - x)}^{2}}$

$f' (x) = \frac{{(1 - x)}^{2} \times [120 x^{11} - 121 x^{10} + 1] + 2 (1 - x) [10 x^{12} - 11 x^{11} + x]}{{(1 - x)}^{4}}$

$Hence, 2 f (2) + f' (2) = 119 \cdot 2^{10} + 1$

$∴$ $n = 10$

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