$$a + ar + ar^2 + … + ar^{n – 1} = \dfrac{a (r^n – 1)}{(r – 1)},$$r ≠ 1

Asked by Aaryan | 1 year ago |  84

##### Solution :-

Let us check for n = 1,

P (1): a =$$\dfrac{ a (r^1 – 1)}{(r-1)}$$

a = a

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): a + ar + ar2 + … + ark – 1 = $$\dfrac{ a (r^k – 1)}{(r – 1)}$$ … (i)

So,

a + ar + ar2 + … + ark – 1 + ark

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

= $$\dfrac{a(rk – 1)}{(r – 1)}$$ + ark by using equation (i)

=$$\dfrac{ a(rk+1 – 1)}{ (r-1)}$$

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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