Let f(x) = sum_{k=1}^{10} kx^k,x in R. If 2f( | vedclass

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(B) Given $f(x) = \sum_{k=1}^{10} kx^k = x + 2x^2 + 3x^3 + \dots + 10x^{10}$.We know that $f'(x) = \sum_{k=1}^{10} k^2 x^{k-1}$.Thus,$f(2) = \sum_{k=1

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$10$

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(B) Given $f(x) = \sum_{k=1}^{10} kx^k = x + 2x^2 + 3x^3 + \dots + 10x^{10}$.We know that $f'(x) = \sum_{k=1}^{10} k^2 x^{k-1}$.Thus,$f(2) = \sum_{k=1}^{10} k(2^k)$ and $f'(2) = \sum_{k=1}^{10} k^2(2^{k-1})$.Consider the identity $g(x) = \sum_{k=1}^{10} x^k = \frac{x(1-x^{10})}{1-x}$.Differentiating $g(x)$,we get $g'(x) = \sum_{k=1}^{10} kx^{k-1} = f(x)/x$.Also,$f(x) = x g'(x)$.Then $f'(x) = g'(x) + x g''(x)$.So,$2f(2) + f'(2) = 2(2g'(2)) + (g'(2) + 2g''(2)) = 5g'(2) + 2g''(2)$.Alternatively,using the property of the sum $\sum k x^k$,we have $f(x) = \frac{x(1-(n+1)x^n + nx^{n+1})}{(1-x)^2}$ for $n=10$.Evaluating at $x=2$,$f(2) = \frac{2(1-11(2^{10}) + 10(2^{11}))}{(1-2)^2} = 2(1 - 11(1024) + 20480) = 2(1 - 11264 + 20480) = 2(9217) = 18434$.Calculating $f'(2)$ and substituting,we find $2f(2) + f'(2) = 119(2^{10}) + 1$.Thus,$n = 10$.

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

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