JEE Mains · Maths · STD 11 - 8. sequence and series
The sum \(1+2 \cdot 3+3 \cdot 3^{2}+\ldots . .+10 \cdot 3^{9}\) is equal to
- A \(\frac{2 \cdot 3^{12}+10}{4}\)
- B \(\frac{19 \cdot 3^{10}+1}{4}\)
- C \(5 \cdot 3^{10}-2\)
- D \(\frac{9 \cdot 3^{10}+1}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{19 \cdot 3^{10}+1}{4}\)
Step-by-step Solution
Detailed explanation
\(S =1 \cdot 3^{0}+2 \cdot 3^{1}+3 \cdot 3^{2}+\ldots . .+10.3^{9}\) \(3 S =1 \cdot 3^{1}+2.3^{2} \ldots \ldots \ldots \ldots \ldots \ldots+9 \times 3^{9}+10 \times 3^{10}\) \(-2 S =\left(1 \cdot 3^{0}+3^{1}+3^{2} \ldots 3^{9}\right)-10.3^{10}\)…
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