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Prove the following by using the principle of mathematical induction for all $n \in N$ \[\] $ \large\frac{1}{1.2.3}$$+\large\frac{1}{2.3.4}$$+\large\frac{1}{3.4.5}$$+...+\large\frac{1}{n(n+1)(n+2)}$$=\large\frac{n(n+3)}{4(n+1)(n+2)}$

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Let the given statement be $P(n)$, i.e.,
$P(n) : \large\frac{1}{1.2.3}$$+\large\frac{1}{2.3.4}$$+\large\frac{1}{3.4.5}$$+...+\large\frac{1}{n(n+1)(n+2)}$$=\large\frac{n(n+3)}{4(n+1)(n+2)}$
For $n = 1$, we have
$P(n) : \large\frac{1}{1.2.3}$$ = \large\frac{1.(1+3)}{4(1+1)(1+2)}$$=\large\frac{1.4}{4.2.3}$$=\large\frac{1}{1.2.3}$, which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$ \large\frac{1}{1.2.3}$$+\large\frac{1}{2.3.4}$$+\large\frac{1}{3.4.5}$$+...+\large\frac{1}{k(k+1)(k+2)}$$=\large\frac{k(k+3)}{4(k+1)(k+2)}$----------(i)
We shall now prove that $P(k+1)$ is true.
$\bigg[ \large\frac{1}{1.2.3}$$+\large\frac{1}{2.3.4}$$+\large\frac{1}{3.4.5}$$+...+\large\frac{1}{k(k+1)(k+2)} \bigg]$$+ \large\frac{1}{(k+1)(k+2)(k+3)}$
$ = \large\frac{k(k+3)}{4(k+1)(k+2)}$$+\large\frac{1}{(k+1)(k+2)(k+3)}$$\qquad$ [ Using (i) ]
$ = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k(k+3)}{4}$$+ \large\frac{1}{(k+3)} \bigg\}$
$ = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k(k+3)^2+4}{4(k+3)}\bigg\}$
$ = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k(k^2+6k+9)+4}{4(k+3)}\bigg\}$
$ = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k^3+6k^2+9k+4}{4(k+3)}\bigg\}$
$ = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k^3+2k^2+k+4k^2+8k+4}{4(k+3)}\bigg\}$
$ = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k(k^2+2k+1)+4(k^2+2k+1)}{4(k+3)}\bigg\}$
$ = \large\frac{1}{(k+1)(k+2)}$$ \bigg\{ \large\frac{k(k+1)^2+4(k+1)^2}{4(k+3)}\bigg\}$
$ = \large\frac{(k+1)^2+(k+4)}{4(k+1)(k+2)(k+3)}$
$ = \large\frac{(k+1) \{ (k+1)+3 \}}{4 \{ (k+1)+1 \}\{ (k+1)+2 \}}$
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., $n$.
answered Apr 30, 2014 by thanvigandhi_1

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