1.3 + 2.4 + 3.5 + … + n. (n+2) =  $$\dfrac{1}{6}$$n (n+1) (2n+7)

Asked by Aaryan | 1 year ago |  102

##### Solution :-

Let us check for n = 1,

P (1): 1.3 = $$\dfrac{1}{6}$$ × 1 × 2 × 9

3 = 3

P (n) is true for n = 1.

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): 1.3 + 2.4 + 3.5 + … + k. (k+2) = $$\dfrac{1}{6}$$ k (k+1) (2k+7) … (i)

So,

1.3 + 2.4 + 3.5 + … + k. (k+2) + (k+1) (k+3)

Now, substituting the value of P (k) we get,

= $$\dfrac{1}{6}$$k (k+1) (2k+7) + (k+1) (k+3) by using equation (i)

=$$(k+1) \dfrac{k(2k+7)}{6} + \dfrac{(k+3)}{1}$$

= $$\dfrac{ (k+1) (2k+9) (k+2) }{ 6}$$

= $$\dfrac{1}{6}$$(k+1) (k+2) (2k+9)

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

Answered by Aaryan | 1 year ago

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