JEE Mains · Maths · STD 11 - 8. sequence and series
The sum of the infinite series \(1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\frac{22}{3^{5}}+\ldots\) is equal to
- A \(\frac{13}{4}\)
- B \(\frac{9}{4}\)
- C \(\frac{15}{4}\)
- D \(\frac{11}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{13}{4}\)
Step-by-step Solution
Detailed explanation
\(S=1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\ldots\) \(\frac{S}{3}=\frac{1}{3}+\frac{2}{3^{2}}+\frac{7}{3^{3}}+\frac{12}{3^{4}}+\ldots\) \(\frac{2 S}{3}=1+\frac{1}{3}+\frac{5}{3^{2}}+\frac{5}{3^{3}}+\frac{5}{3^{4}}+\ldots+\) up to infinite terms…
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