TS EAMCET · Maths · Sequences and Series
If \(x=1+\frac{3}{1 !} \times \frac{1}{6}+\frac{3 \times 7}{2 !}\left(\frac{1}{6}\right)^2\) \(+\frac{3 \times 7 \times 11}{3 !}\left(\frac{1}{6}\right)^3+\ldots\), then \(x^4\) equals
- A 81
- B 54
- C 27
- D 8
Answer & Solution
Correct Answer
(C) 27
Step-by-step Solution
Detailed explanation
\begin{aligned} & \because x=1+\frac{3}{1 !} \times \frac{1}{6}+\frac{3 \times 7}{2 !}\left(\frac{1}{6}\right)^2 \\ &+\frac{3 \times 7 \times 11}{3 !}\left(\frac{1}{6}\right)^3+\ldots \ldots \\ & \therefore \quad(1-\alpha)^{-p / q} \\ &= 1+\left(\frac{p}{q}\right) \frac{1}{1…
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