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Question: Answered & Verified by Expert
$\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots$ upto $n$ terms is equal to
MathematicsSequences and SeriesCOMEDKCOMEDK 2019
Options:
  • A $\frac{n}{4 n+6}$
  • B $\frac{1}{4 n+4}$
  • C $\frac{n}{6 n+4}$
  • D $\frac{n}{3 n+7}$
Solution:
1058 Upvotes Verified Answer
The correct answer is: $\frac{n}{6 n+4}$
We have,
$$
\begin{aligned}
&=\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{[(2+(n-1) 3)][5+(n-1) 3]} \\
&=\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{(3 n-1)(3 n+2)} \\
&=\frac{1}{3}\left[\frac{3}{2 \cdot 5}+\frac{3}{5 \cdot 8}+\frac{3}{8 \cdot 11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}\right] \\
&=\frac{1}{3}\left[\frac{5-2}{2 \cdot 5}+\frac{8-5}{5 \cdot 8}+\frac{11-8}{8 \cdot 11}+\ldots+\frac{(3 n+2)-(3 n-1)}{(3 n-1)(3 n+2)}\right] \\
&=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\ldots+\frac{1}{3 n-1}-\frac{1}{3 n+2}\right] \\
&=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3 n+2}\right]=\frac{1}{3}\left[\frac{3 n+2-2}{2(3 n+2)}\right]=\frac{n}{6 n+4}
\end{aligned}
$$

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