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JEE Mains · Maths · STD 11 - 8. sequence and series

The greatest integer less than or equal to the sum of first \(100\) terms of the sequence \(\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, \ldots \ldots\) is equal to

  1. A \(99\)
  2. B \(98\)
  3. C \(89\)
  4. D \(88\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(98\)

Step-by-step Solution

Detailed explanation

\(\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}+\ldots\) \(\left(1-\frac{2}{3}\right)+\left(1-\frac{4}{9}\right)+\left(1-\frac{8}{27}\right)+\left(1-\frac{16}{81}\right) \ldots .100 \text { terms }\) \(100-\left[\frac{2}{3}+\left(\frac{2}{3}\right)^{2}+\ldots\right]\)…
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