AP EAMCET · Maths · Binomial Theorem
\(1+\frac{4}{15}+\frac{4.10}{15.30}+\frac{4.10 .16}{15.30 .45}+\ldots \quad \infty=\)
- A \(\left(\frac{3}{5}\right)^{2 / 3}\)
- B \(\left(\frac{5}{3}\right)^{2 / 3}\)
- C \(\left(\frac{3}{5}\right)^{3 / 2}\)
- D \(\left(\frac{5}{3}\right)^{3 / 2}\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{5}{3}\right)^{2 / 3}\)
Step-by-step Solution
Detailed explanation
\(ny = \frac{4}{15}\), \(\frac{n(n-1)}{2!}y^2 = \frac{4 \cdot 10}{15 \cdot 30} = \frac{4}{45}\) \(\frac{n-1}{2n} = \frac{4/45}{(4/15)^2} = \frac{5}{4} \implies 4(n-1)=10n \implies 6n=-4 \implies n=-\frac{2}{3}\) \((-\frac{2}{3})y = \frac{4}{15} \implies y = -\frac{2}{5}\)…
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