Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): \frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$

For n = 1, we have

$P(1): \frac{1}{1 \cdot 2 \cdot 3}=\frac{1 \cdot(1+3)}{4(1+1)(1+2)}=\frac{1 \cdot 4}{4 \cdot 2 \cdot 3}=\frac{1}{1 \cdot 2 \cdot 3}$, which is true.

Let P(k) be true for some positive integer k, i.e.,

$\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)}$ (i)

We shall now prove that P(k + 1) is true.

Consider

$\left[\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots . .+\frac{1}{k(k+1)(k+2)}\right]+\frac{1}{(k+1)(k+2)(k+3)}$

$=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$ [Using (i)]

$=\frac{1}{(k+1)(k+2)}\left\{\frac{k(k+3)}{4}+\frac{1}{k+3}\right\}$

$=\frac{1}{(k+1)(k+2)}\left\{\frac{k(k+3)^{2}+4}{4(k+3)}\right\}$

$=\frac{1}{(k+1)(k+2)}\left\{\frac{k\left(k^{2}+6 k+9\right)+4}{4(k+3)}\right\}$

$=\frac{1}{(k+1)(k+2)}\left\{\frac{k^{3}+6 k^{2}+9 k+4}{4(k+3)}\right\}$

$=\frac{1}{(k+1)(k+2)}\left\{\frac{k^{3}+2 k^{2}+k+4 k^{2}+8 k+4}{4(k+3)}\right\}$

$=\frac{1}{(k+1)(k+2)}\left\{\frac{k(k+1)^{2}+4(k+1)^{2}}{4(k+3)}\right\}$

$=\frac{(k+1)^{2}(k+4)}{4(k+1)(k+2)(k+3)}$

$=\frac{(k+1)\{(k+1)+3\}}{4\{(k+1)+1\}\{(k+1)+2\}}$

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.