2 + 5 + 8 + 11 + … + (3n – 1) = $$\dfrac{1}{2}$$n (3n + 1)

Asked by Aaryan | 1 year ago |  67

##### Solution :-

Let us check for n = 1,

P (1): 2 = $$\dfrac{1}{2}$$ × 1 × 4

2 = 2

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) = 2 + 5 + 8 + 11 + … + (3k – 1) =$$\dfrac{1}{2}$$ k (3k + 1) … (i)

So,

2 + 5 + 8 + 11 + … + (3k – 1) + (3k + 2)

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

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

= $$\dfrac{3k^2 + 7k + 2} { 2}$$

= $$\dfrac{3k^2 + 4k + 3k + 2 }{ 2}$$

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

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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