TS EAMCET · Maths · Sequences and Series
\(1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots\) is equal to :
- A \(4^{-2 / 3}\)
- B \(\sqrt[3]{16}\)
- C \(\sqrt[3]{4}\)
- D \(4^{3 / 2}\)
Answer & Solution
Correct Answer
(B) \(\sqrt[3]{16}\)
Step-by-step Solution
Detailed explanation
Let \(S=1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\ldots\) On comparing with \((1+x)^n=1+n x+\frac{n(n-1)}{2 !} x^2+\ldots\) we get \(n x=\frac{2}{4}\) \(\ldots\) (i) and \(\quad \frac{n(n-1)}{2 !} x^2=\frac{2 \cdot 5}{4 \cdot 8}\)…
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