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

Option: 1

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Option: 2

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Option: 3

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Option: 4

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Answers (1)

best_answer

10=1+\frac{4}{\mathrm{k}}+\frac{8}{\mathrm{k}^{2}}+\frac{13}{\mathrm{k}^{3}}+\frac{19}{\mathrm{k}^{4}}+\ldots .$. upto $\infty
9=\frac{4}{\mathrm{k}}+\frac{8}{\mathrm{k}^{2}}+\frac{13}{\mathrm{k}^{3}}+\frac{19}{\mathrm{k}^{4}}+\ldots .$. upto $\infty
\frac{9}{\mathrm{k}}=\frac{4}{\mathrm{k}^{2}}+\frac{8}{\mathrm{k}^{3}}+\frac{13}{\mathrm{k}^{4}}+\ldots .$. upto $\infty
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\mathrm{S}=9\left(1-\frac{1}{\mathrm{k}}\right)=\frac{4}{\mathrm{k}}+\frac{4}{\mathrm{k}^{2}}+\frac{5}{\mathrm{k}^{3}}+\frac{6}{\mathrm{k}^{4}} \ldots .$. upto $\infty
{\frac{\mathrm{S}}{\mathrm{k}}}=\frac{4}{\mathrm{k}^{2}}+\frac{4}{\mathrm{k}^{3}}+\frac{5}{\mathrm{k}^{4}}+\ldots .$. upto $\infty
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\left(1-\frac{1}{\mathrm{k}}\right) \mathrm{S}=\frac{4}{\mathrm{k}}+\frac{1}{\mathrm{k}^{3}}+\frac{1}{\mathrm{k}^{4}}+\frac{1}{\mathrm{k}^{5}}+\ldots . . \infty

9\left(1-\frac{1}{\mathrm{k}}\right)^{2}=\frac{4}{\mathrm{k}}+\frac{\frac{1}{\mathrm{k}^{3}}}{\left(1-\frac{1}{\mathrm{k}}\right)}
9(\mathrm{k}-1)^{3}=4 \mathrm{k}(\mathrm{k}-1)+1
k=2

Posted by

SANGALDEEP SINGH

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