For k , if the sum of the series 1 + 4 k + 8 k 2 + 13 k 3 + 19 k 4 + . . . is 10, then the value of k is __________ . [2023]

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Solution

Q.
For $k \in ℕ$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k^{2}} + \frac{13}{k^{3}} + \frac{19}{k^{4}} + . . .$ is 10, then the value of k is __________ . [2023]

Ans.
(2)

Given, $10 = 1 + \frac{4}{k} + \frac{8}{k^{2}} + \frac{13}{k^{3}} + \frac{19}{k^{4}} + \dots up to \infty$

$\Rightarrow 9 = \frac{4}{k} + \frac{8}{k^{2}} + \frac{13}{k^{3}} + \frac{19}{k^{4}} + \dots up to \infty$ ...(i)

$\Rightarrow \frac{9}{k} = \frac{4}{k^{2}} + \frac{8}{k^{3}} + \frac{13}{k^{4}} + \frac{19}{k^{5}} + \dots up to \infty$ ...(ii)

Subtract (ii) from (i), we get

$S = 9 (1 - \frac{1}{k}) = \frac{4}{k} + \frac{4}{k^{2}} + \frac{5}{k^{3}} + \frac{6}{k^{4}} + \dots up to \infty$ ...(iii)

$\frac{S}{k} = \frac{4}{k^{2}} + \frac{4}{k^{3}} + \frac{5}{k^{4}} + \frac{6}{k^{5}} + \dots up to \infty$ ...(iv)

Subtract (iv) from (iii), we get

$(1 - \frac{1}{k}) S = \frac{4}{k} + \frac{1}{k^{3}} + \frac{1}{k^{4}} + \frac{1}{k^{5}} + \dots up to \infty$ ...(v)

Putting $S = 9 (1 - \frac{1}{k})$ from (iii) in (v), we get

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

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

$\Rightarrow 9 {(k - 1)}^{3} = 4 k (k - 1) + 1$

Put $k - 1 = x$ , then

$9 x^{3} = 4 (x + 1) x + 1 \Rightarrow 9 x^{3} = 4 x^{2} + 4 x + 1$

$\Rightarrow 9 x^{3} - 4 x^{2} - 4 x - 1 = 0$

$\Rightarrow (x - 1) (9 x^{2} + 5 x + 1) = 0 \Rightarrow (x - 1) = 0 \Rightarrow x = 1$

$∴$ $k - 1 = 1 \Rightarrow k = 2$

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