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$\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+$ equal to
upto $n$ terms is
Options:
upto $n$ terms is
Solution:
1881 Upvotes
Verified Answer
The correct answer is:
$\frac{n}{6 n+4}$
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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