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

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Sequences and Series

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}} + \dots

is 10, then the value of k is

Answer: 2

Write

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

, where the numerators

1, 4, 8, 13, 19, \dots

have first differences

3, 4, 5, 6, \dots

. Multiplying by

\frac{1}{k}

and subtracting,

(1 - \frac{1}{k}) S = 1 + \frac{3}{k} + \frac{4}{k ^{2}} + \frac{5}{k ^{3}} + \dots

, whose numerators now increase by a constant. Repeating the step,

(1 - \frac{1}{k})^{2} S = 1 + \frac{2}{k} + \frac{1}{k ^{2}} + \frac{1}{k ^{3}} + \dots = 1 + \frac{2}{k} + \frac{1}{k ( k - 1 )}

, since the tail is a geometric series. With

(1 - \frac{1}{k})^{2} = \frac{( k - 1 ) ^{2}}{k ^{2}}

and

S = 10

, clearing denominators gives

10 (k - 1)^{3} = k^{3} + k^{2} - k

, i.e.

9 k^{3} - 31 k^{2} + 31 k - 10 = 0

. This factors as

(k - 2) (9 k^{2} - 13 k + 5) = 0

, and the quadratic factor has negative discriminant, so the value of

k

2

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Integer answer