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Question -

a + ar + ar2 + … +arn – 1 = a [(rn – 1)/(r – 1)], r ≠ 1



Answer -

Let P (n): a + ar + ar2 + … + arn– 1 = a [(rn – 1)/(r – 1)]

Let us check for n = 1,

P (1): a = a (r1 – 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 toprove that P (k + 1) is true.

P (k): a + ar + ar2 + … + ark – 1 =a [(rk – 1)/(r – 1)] … (i)

So,

a + ar + ar2 + … + ark – 1 +ark

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

= a [(rk – 1)/(r – 1)] + ark byusing equation (i)

= a[rk – 1 + rk(r-1)] / (r-1)

= a[rk – 1 + rk+1 – r‑k]/ (r-1)

= a[rk+1 – 1] / (r-1)

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

Hence, P (n) is true for all n N.

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