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$\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots \frac{1}{(3 n-1)(3 n+2)}$ is equal to
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Verified Answer
The correct answer is:
$\frac{n}{6 n+4}$
Given, $\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots$
$$
\begin{array}{r}
5 \cdot 8 \quad 8 \cdot 11 \\
=\frac{1}{3}\left[\left(\frac{1}{2}-\frac{1}{5}\right)+\frac{1}{(3 n-1)(3 n+2)}\right. \\
+\ldots+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right) \\
\left.\left.+\ldots-\frac{1}{3 n+2}\right)\right]
\end{array}
$$
$$
\begin{aligned}
&=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3 n+2}\right] \\
&=\frac{3 n+2-2}{6(3 n+2)}=\frac{n}{6 n+4}
\end{aligned}
$$
$$
\begin{array}{r}
5 \cdot 8 \quad 8 \cdot 11 \\
=\frac{1}{3}\left[\left(\frac{1}{2}-\frac{1}{5}\right)+\frac{1}{(3 n-1)(3 n+2)}\right. \\
+\ldots+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right) \\
\left.\left.+\ldots-\frac{1}{3 n+2}\right)\right]
\end{array}
$$
$$
\begin{aligned}
&=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3 n+2}\right] \\
&=\frac{3 n+2-2}{6(3 n+2)}=\frac{n}{6 n+4}
\end{aligned}
$$
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