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

Asked by Aaryan | 2 years ago |  118

1 Answer

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 | 2 years ago

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