JEE Mains · Maths · STD 11 - 8. sequence and series
If \(\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^{9}}+\ldots . .+\frac{10240}{3}=2^{ n } \cdot m\), where \(m\) is odd, then \(m . n\) is equal to
- A \(15\)
- B \(14\)
- C \(13\)
- D \(12\)
Answer & Solution
Correct Answer
(D) \(12\)
Step-by-step Solution
Detailed explanation
\(\frac{6}{3^{12}}+10\left(\frac{1}{3^{11}}+\frac{2}{3^{10}}+\frac{2^{2}}{3^{9}}+\frac{2^{3}}{3^{8}}+\ldots .+\frac{2^{10}}{3}\right)\) \(\frac{6}{3^{12}}+\frac{10}{3^{11}}\left(\frac{6^{11}-1}{6-1}\right)\) \(=2^{12} \cdot 1 ; m . n =12\)
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