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Question: 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 _____...

Let f(x)=∑k=110kxk,x∈Rf(x) = \sum_{k=1}^{10} kx^k, x \in R. If 2f(2)+f′(2)=119(2)n+12f(2)+f'(2)=119(2)^n+1 then n is equal to ______

Answer

10

Explanation

Solution

Let G(x)=∑k=010xk=x11−1x−1G(x) = \sum_{k=0}^{10} x^k = \frac{x^{11}-1}{x-1}. Then f(x)=xG′(x)f(x) = x G'(x). G′(x)=11x10(x−1)−(x11−1)(x−1)2=10x11−11x10+1(x−1)2G'(x) = \frac{11x^{10}(x-1) - (x^{11}-1)}{(x-1)^2} = \frac{10x^{11}-11x^{10}+1}{(x-1)^2}. At x=2x=2, G′(2)=10(2)11−11(2)10+1(2−1)2=20(2)10−11(2)10+1=9(2)10+1G'(2) = \frac{10(2)^{11}-11(2)^{10}+1}{(2-1)^2} = 20(2)^{10}-11(2)^{10}+1 = 9(2)^{10}+1. f(2)=2G′(2)=2(9(2)10+1)=9(2)11+2f(2) = 2 G'(2) = 2(9(2)^{10}+1) = 9(2)^{11}+2. f′(x)=G′(x)+xG′′(x)f'(x) = G'(x) + x G''(x). G′′(x)=ddx(10x11−11x10+1(x−1)2)G''(x) = \frac{d}{dx} \left( \frac{10x^{11}-11x^{10}+1}{(x-1)^2} \right). At x=2x=2, G′′(2)=(110⋅210−110⋅29)(1)−2(10⋅211−11⋅210+1)(1)2=110⋅29−2(9⋅210+1)=55⋅210−18⋅210−2=37⋅210−2G''(2) = \frac{(110 \cdot 2^{10} - 110 \cdot 2^9)(1) - 2(10 \cdot 2^{11} - 11 \cdot 2^{10} + 1)}{(1)^2} = 110 \cdot 2^9 - 2(9 \cdot 2^{10} + 1) = 55 \cdot 2^{10} - 18 \cdot 2^{10} - 2 = 37 \cdot 2^{10} - 2. f′(2)=G′(2)+2G′′(2)=(9⋅210+1)+2(37⋅210−2)=9⋅210+1+74⋅210−4=83⋅210−3f'(2) = G'(2) + 2 G''(2) = (9 \cdot 2^{10} + 1) + 2(37 \cdot 2^{10} - 2) = 9 \cdot 2^{10} + 1 + 74 \cdot 2^{10} - 4 = 83 \cdot 2^{10} - 3. 2f(2)+f′(2)=2(9⋅211+2)+(83⋅210−3)=9⋅212+4+83⋅210−3=36⋅210+83⋅210+1=119⋅210+12f(2) + f'(2) = 2(9 \cdot 2^{11} + 2) + (83 \cdot 2^{10} - 3) = 9 \cdot 2^{12} + 4 + 83 \cdot 2^{10} - 3 = 36 \cdot 2^{10} + 83 \cdot 2^{10} + 1 = 119 \cdot 2^{10} + 1. Comparing with 119(2)n+1119(2)^n+1, we get n=10n=10.