TS EAMCET · Maths · Sequences and Series
If \(\frac{1}{2 \times 4}+\frac{1}{4 \times 6}+\frac{1}{6 \times 8}+\ldots(n\) terms \()=\frac{k n}{n \times 1}\), then \(k\) is equal to
- A \(\frac{1}{4}\)
- B \(\frac{1}{2}\)
- C \(1\)
- D \(\frac{1}{8}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
Given series can be rewritten as \(\begin{gathered}\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots+\frac{1}{2 n}-\frac{1}{2 n+2}\right] \\ \quad=\frac{1}{4}\left(1-\frac{1}{n+1}\right) \\ =\frac{1}{4}\left(\frac{n}{n+1}\right)\end{gathered}\) On comparing…
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