For k N, if the sum of the series 1+(4)/(k)+8k^2+13k^3+19k^4+ is 10 , then the value of k …

Question

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 \_\_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

From the given series :

$10=1+\frac{4}{k}+\frac{8}{{k}^{2}}+\frac{13}{{k}^{3}}+\frac{19}{{k}^{4}}+\ldots$

We isolate the 1 to get :

$9=\frac{4}{k}+\frac{8}{{k}^{2}}+\frac{13}{{k}^{3}}+\frac{19}{{k}^{4}}+\ldots$ .......(1)

Divide each term in the equation by $k$ :

$\frac{9}{k}=\frac{4}{{k}^{2}}+\frac{8}{{k}^{3}}+\frac{13}{{k}^{4}}+\ldots$ .......(2)

Subtracting the second equation from the first, we obtain :

$9(1-\frac{1}{k})=\frac{4}{k}+\frac{4}{{k}^{2}}+\frac{5}{{k}^{3}}+\frac{6}{{k}^{4}}+\ldots$

This is equivalent to :

$S=9(1-\frac{1}{k})=\frac{4}{k}+\frac{4}{{k}^{2}}+\frac{5}{{k}^{3}}+\frac{6}{{k}^{4}}+\ldots$ .........(3)

Divide both sides by $k$ again, we get :
$\frac{S}{k}=\frac{4}{{k}^{2}}+\frac{4}{{k}^{3}}+\frac{5}{{k}^{4}}+\ldots$ ..........(4)

Subtracting equation (4) from (3), we have :

$(1-\frac{1}{k})S=\frac{4}{k}+\frac{1}{{k}^{3}}+\frac{1}{{k}^{4}}+\ldots$

In this equation, you've treated the infinite series on the right-hand side as a geometric series with a ratio of $1/k$, so the sum of this series can be expressed as $\frac{1/{k}^{3}}{1-1/k}$.

Therefore, we get :

$9{(1-\frac{1}{k})}^{2}=\frac{4}{k}+\frac{1/{k}^{3}}{1-1/k}.$

Now, simplifying this equation, we get :

$9(k-1{)}^{2}=4{k}^{2}-4k+1$

$\Rightarrow$$9({k}^{2}-2k+1)=4{k}^{2}-4k+1$

$\Rightarrow$$9{k}^{2}-18k+9=4{k}^{2}-4k+1$

$\Rightarrow$$9{k}^{2}-4{k}^{2}-18k+4k+9-1=0$

$\Rightarrow$$5{k}^{2}-14k+8=0$

This is a standard form quadratic equation, $a{x}^{2}+bx+c=0$, where $a=5$, $b=-14$, and $c=8$.

The solutions can be found using the quadratic formula, $k=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$.

Substituting the values for $a$, $b$, and $c$, we have :

$\Rightarrow$$k=\frac{14\pm \sqrt{(-14{)}^{2}-4\times 5\times 8}}{2\times 5}$

$\Rightarrow$$k=\frac{14\pm \sqrt{196-160}}{10}$

$\Rightarrow$$k=\frac{14\pm \sqrt{36}}{10}$

$\Rightarrow$$k=\frac{14\pm 6}{10}$

So, the solutions are $k=2$ or $k=0.8$. However, since the question mentions $k\in N$, i.e., $k$ is a natural number, the only valid solution is $k=2$.

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