KCET · Maths · Sequences and Series
\(\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\) equal to
upto \(n\) terms is
- A \(\frac{n}{4 n+6}\)
- B \(\frac{1}{6 n+4}\)
- C \(\frac{n}{6 n+4}\)
- D \(\frac{n}{3 n+7}\)
Answer & Solution
Correct Answer
(C) \(\frac{n}{6 n+4}\)
Step-by-step Solution
Detailed explanation
Let
\[
\begin{aligned}
\mathrm{S}_{\mathrm{n}} &=\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{(3 n-1)(3 n+2)} \\
&=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\ldots \ldots+\frac{1}{3 n-1}-\frac{1}{3 n+2}\right] \\
&=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3 n+2}\right] \\
&=\frac{n}{6 n+4}
\end{aligned}
\]
\[
\begin{aligned}
\mathrm{S}_{\mathrm{n}} &=\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{(3 n-1)(3 n+2)} \\
&=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\ldots \ldots+\frac{1}{3 n-1}-\frac{1}{3 n+2}\right] \\
&=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3 n+2}\right] \\
&=\frac{n}{6 n+4}
\end{aligned}
\]
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