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

$$1.2+2.3+3.4+...+n(n+1)$$

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

Asked by Abhisek | 1 year ago |  52

##### Solution :-

$$1.2+2.3+3.4+...+n(n+1) =[\dfrac{n(n+1)(n+2)}{3}]$$

For n=1,

L.H.S.=1.2=2

R.H.S.  $$=\dfrac{1(1+1)(1+2)}{3}$$

$$\dfrac{1.2.3}{3}=2$$

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

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

$$1.2+2.3+3.4+...+k(k+1)$$

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

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

Consider

$$1.2+2.3+3.4+...+k(k+1)$$

$$+(k+1)(k+2)$$

$$=\dfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)$$ [using (i)]

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

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

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

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 Abhisek | 1 year ago

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