JEE Mains · Maths · STD 11 - 8. sequence and series
The sum of the infinite series \(1+\frac{5}{6}+\frac{12}{6^{2}}+\frac{22}{6^{3}}+\frac{35}{6^{4}}+\frac{51}{6^{5}}+\frac{70}{6^{6}}+\ldots .\) is equal to
- A \(\frac{425}{216}\)
- B \(\frac{429}{216}\)
- C \(\frac{288}{125}\)
- D \(\frac{280}{125}\)
Answer & Solution
Correct Answer
(C) \(\frac{288}{125}\)
Step-by-step Solution
Detailed explanation
\(S=1+\frac{5}{6}+\frac{12}{6^{2}}+\frac{22}{6^{3}}+\frac{35}{6^{4}}+\ldots\) \(\frac{S}{6}=\frac{1}{6}+\frac{5}{6^{2}}+\frac{12}{6^{3}}+\frac{22}{6^{4}}+\ldots\) on subtraction \(\frac{5}{6} S=1+\frac{4}{6}+\frac{7}{6^{2}}+\frac{10}{6^{3}}+\frac{13}{6^{4}}+\ldots\)…
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