JEE Mains · Maths · STD 11 - 8. sequence and series
If \(2^{10}+2^{9} \cdot 3^{1}+28 \cdot 3^{2}+\ldots+2 \cdot 3^{9}+3^{10}=S -211\) then \(S\) is equal to
- A \(\frac{3^{11}}{2}+2^{10}\)
- B \(3^{11}-2^{12}\)
- C \(3^{11}\)
- D \(2 \cdot 3^{11}\)
Answer & Solution
Correct Answer
(C) \(3^{11}\)
Step-by-step Solution
Detailed explanation
\(a =2^{10} ; r =\frac{3}{2} ; n =11( G \cdot P )\) \(S^{\prime}=\left(2^{10}\right) \frac{\left(\left(\frac{3}{2}\right)^{11}-1\right)}{\frac{3}{2}-1}=2^{11}\left(\frac{3^{11}}{2^{11}}-1\right)\) \(S^{\prime}=3^{11}-2^{11}= S -2^{11}( Given )\) \(\therefore S =3^{11}\)
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