JEE Mains · Maths · STD 11 - 8. sequence and series
\(\frac{6}{3^{26}}+\frac{10 \cdot 1}{3^{25}}+\frac{10 \cdot 2}{3^{24}}+\frac{10 \cdot 2^2}{3^{23}}+\ldots+\frac{10 \cdot 2^{24}}{3}\) is equal to
- A \(2^{25}\)
- B \(2^{26}\)
- C \(3^{25}\)
- D \(3^{26}\)
Answer & Solution
Correct Answer
(B) \(2^{26}\)
Step-by-step Solution
Detailed explanation
\(S=\frac{6}{3^{26}}+\frac{10}{3^{25}}[\frac{(6)^{25}-1}{6-1}]\) \(S=\frac{6}{3^{26}}+\frac{10}{3^{25}}[\frac{6^{25}-1}{5}]\) \(S =\frac{2}{3^{25}}+2\left[2^{25}-\frac{1}{3^{25}}\right]\) \(S=2^{26}\)
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