For k in N,if the sum of the series 1+frac{4} | vedclass

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(A) Let $S = 1 + \frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots = 10$. Subtracting $1$ from both sides,we get $\frac{4}{k} + \

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$2$

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(A) Let $S = 1 + \frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots = 10$. Subtracting $1$ from both sides,we get $\frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots = 9$. Let $S_1 = \frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots = 9$. Then $\frac{S_1}{k} = \frac{4}{k^2} + \frac{8}{k^3} + \frac{13}{k^4} + \ldots$. Subtracting these: $S_1(1 - \frac{1}{k}) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots = 9(1 - \frac{1}{k})$. Let $S_2 = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$. Then $\frac{S_2}{k} = \frac{4}{k^2} + \frac{4}{k^3} + \frac{5}{k^4} + \ldots$. Subtracting these: $S_2(1 - \frac{1}{k}) = \frac{4}{k} + \frac{1}{k^3} + \frac{1}{k^4} + \ldots = \frac{4}{k} + \frac{1/k^3}{1 - 1/k} = \frac{4}{k} + \frac{1}{k^2(k-1)}$. Substituting $S_2 = 9(1 - \frac{1}{k})$,we get $9(1 - \frac{1}{k})^2 = \frac{4}{k} + \frac{1}{k^2(k-1)}$. $9(\frac{k-1}{k})^2 = \frac{4k(k-1) + 1}{k^2(k-1)}$. $9(k-1)^3 = 4k^2 - 4k + 1 = (2k-1)^2$. Testing $k=2$: $9(2-1)^3 = 9(1) = 9$,and $(2(2)-1)^2 = 3^2 = 9$. Thus,$k=2$ is the solution.

For $k \in N$,if the sum of the series $1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$ is $10$,then the value of $k$ is

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