JEE Mains · Maths · STD 11 - 8. sequence and series
If the value of \(\left(1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\ldots \text { upto } \infty\right)^{\log _{(0.25)}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . \text { uptow }\right)}\) is \(l\), then \(l^{2}\) is equal to \(......\)
- A \(2\)
- B \(3\)
- C \(4\)
- D \(5\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(\ell=(\underbrace{1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}}_{\mathrm{s}}+\ldots)^{\log _{(0.25)\left(\frac{1}{3}+\frac{1}{3^{2}}+\ldots\right)}}\) \(S=1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\ldots\)…
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