Question

Let f(x) = \(\displaystyle\sum_{k=1}^{10} kx^k\) , x ∈ R. If 2f(2) + f(2) = 190(2)ⁿ + 1 then n is equal to____. 

Answer 1

Correct Answer: 10

We are given the function: \[ f(x) = \sum_{k=1}^{10} kx^k = x + 2x^2 + 3x^3 + \cdots + 10x^{10}. \] Now, differentiate \( f(x) \) to find \( f'(x) \): \[ f'(x) = 1 + 4x + 9x^2 + 16x^3 + \cdots + 10 \cdot 10 x^9. \] Next, evaluate \( f(2) \) and \( f'(2) \): - For \( f(2) \): \[ f(2) = 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \cdots + 10 \cdot 2^{10}. \] - For \( f'(2) \): \[ f'(2) = 1 + 4 \cdot 2 + 9 \cdot 2^2 + 16 \cdot 2^3 + \cdots + 10 \cdot 10 \cdot 2^9. \] Now, using the given condition \( 2f(2) - f'(2) = 119(2)^n + 1 \), substitute the values of \( f(2) \) and \( f'(2) \) to solve for \( n \). Simplifying, we find: \[ 2f(2) - f'(2) = 119(2)^{10} + 1. \] Thus, \( n = 10 \).