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# Prove the following by using the principle of mathematical induction for all $n \in N$  $\large\frac{1}{1.4}$$+\large\frac{1}{4.7}$$+\large\frac{1}{7.10}$$+...+\large\frac{1}{(3n-2)(3n+1)}$$=\large\frac{n}{(3n+1)}$

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A)
Let the given statement be $P(n)$, i.e.,
$P(n) :\large\frac{1}{1.4}$$+\large\frac{1}{4.7}$$+\large\frac{1}{7.10}$$+...+\large\frac{1}{(3n-2)(3n+1)}$$=\large\frac{n}{(3n+1)}$
For $n = 1$, we have
$P(1) =\large\frac{1}{1.4}$$=\large\frac{1}{3.1+1}$$=\large\frac{1}{4}$$=\large\frac{1}{1.4}, which is true. Let P(k) be true for some positive integer k, i.e., P(k) :\large\frac{1}{1.4}$$+\large\frac{1}{4.7}$$+\large\frac{1}{7.10}$$+...+\large\frac{1}{(3k-2)(3k+1)}$$=\large\frac{k}{(3k+1)}---------(1) We shall now prove that P(k+1) is true. Consider \bigg\{ \large\frac{1}{1.4}$$+\large\frac{1}{4.7}$$+\large\frac{1}{7.10}$$+...+\large\frac{1}{(3k-2)(3k+1)}\bigg\}$$+ \large\frac{1}{\{3(k+1)-2\}\{3(k+1)+1\}} = \large\frac{k}{3k+1}$$ +\large\frac{1}{(3k+1)(3k+4)}\qquad$ [ Using (1) ]
$= \large\frac{1}{(3k+1)}$$\bigg\{ k+\large\frac{1}{(3k+4)} \bigg\} = \large\frac{1}{(3k+1)}$$\bigg\{ \large\frac{k(3k+4)+1}{(3k+4)} \bigg\}$
$= \large\frac{1}{(3k+1)}$$\bigg\{ \large\frac{3k^2+4k+1}{(3k+4)} \bigg\} = \large\frac{1}{(3k+1)}$$\bigg\{ \large\frac{3k^2+3k+k+1}{(3k+4)} \bigg\}$
$= \large\frac{(3k+1)(k+1)}{(3k+1)(3k+4)}$
$= \large\frac{(k+1)}{3(k+1)+1}$
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.