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

To solve for \(n\) in the equation given by \(2f(2) + f(2) = 190(2)^n + 1\), where \(f(x) = \sum_{k=1}^{10} kx^k\), we begin by computing \(f(2)\).

First, calculate each term in the sum:

• \(f(2) = 1(2^1) + 2(2^2) + 3(2^3) + \ldots + 10(2^{10})\)

This expands to:

• \(2 + 8 + 24 + 64 + 160 + 384 + 896 + 2048 + 4608 + 10240\) 

Summing these values gives \(f(2) = 18690\).

Substitute \(f(2)\) into the original equation:

• \(3f(2) = 2 \times 18690 + 18690 = 56070\)

Set this equal to the given expression on the right:

• \(56070 = 190(2)^n + 1\)

Rearrange to solve for \(2^n\):

• \(190(2)^n = 56069\)

Calculate \(2^n\) as:

• \(2^n = \frac{56069}{190}\)

Simplifying gives:

• \(2^n \approx 295.1\)

Calculate \(n\) by recognizing that \(2^8 = 256\) and \(2^9 = 512\). Since \(295.1\) is closer to \(512\), check the exact value: \(2^8 \leq 2^n < 2^9\), so \(n = 8\).

The final answer for \(n\) is \(n = 8\), meeting the range requirement of 10,10.