JEE Mains · Maths · STD 11 - 8. sequence and series
For \(k \in N\), if the sum of the series \(1+\frac{4}{ k }+\frac{8}{ k ^2}+\frac{13}{ k ^3}+\frac{19}{ k ^4}+\ldots\) is 10 , then the value of \(k\) is
- A \(2\)
- B \(4\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(10=1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots\). upto \(\infty\) \(9=\frac{4}{ k }+\frac{8}{ k ^2}+\frac{13}{ k ^3}+\frac{19}{ k ^4}+\ldots \text { upto } \infty\)…
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