Prove the following by using the principle of mathematical induction for all $$n\in N$$

$$\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...$$

$$+\dfrac{1}{n(n+1)(n+2)}=\dfrac{n(n+3)}{4(n+1)(n+2)}$$

Asked by Pragya Singh | 1 year ago |  69

##### Solution :-

Let us denote the given equality by P(n), i.e.,

$$P(n): \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....$$

$$+\dfrac{1}{n(n+1)(n+2)}=\dfrac{n(n+3)}{4(n+1)(n+2)}$$

For n=1,

L.H.S. = $$\dfrac{1}{1.2.3}=\dfrac{1}{6}$$

R.H.S.= $$\dfrac{1.(1+3)}{4(1+1)(1+2)}$$

$$\dfrac{1.4}{4.2.3}=\dfrac{1}{6}$$

Therefore, P(n) is true for n=1.

Let us assume that P(k) is true for some positive integer k , i.e.,

$$\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....$$

$$+\dfrac{1}{k(k+1)(k+2)}=\dfrac{k(k+3)}{4(k+1)(k+2)}$$..........(i)

Now, we have to prove that P(k+1) is also true.

Consider

$$[ \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....$$

$$+\dfrac{1}{k(k+1)(k+2)}]+\dfrac{1}{(k+1)(k+2)(k+3)}$$

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

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

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

$$\dfrac{1}{(k+1)(k+2)}\{\dfrac{k^3+6k^2+9k+4}{4(k+3)}\}$$

$$\dfrac{1}{(k+1)(k+2)}\{\dfrac{k^3+2k^2+k+4k^2+8k+4}{4(k+3)}\}$$

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

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

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

Therefore, P(k+1) holds whenever P(k) holds.

Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction.

Answered by Pragya Singh | 1 year ago

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