Prove by the principle of mathematical induction: a + ar + ar2 + …

1.	Prove by the principle of mathematical induction: a + ar + ar2 + … + arn – 1 = a[(rn – 1)/(r – 1)], r ≠ 1
Answer» Suppose P (n): a + ar + ar² + … + ar^{n – 1} = a[(rⁿ – 1)/(r – 1)] Now let us check for n = 1, P (1): a = a (r¹ – 1)/(r - 1) : a = a P (n) is true for n = 1. Then, 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 + … + ar^{k – 1} = a[(r^k – 1)/(r – 1)] … (i) Therefore, a + ar + ar² + … + ar^{k – 1} + ar^k Then, substituting the value of P (k) we get, = a[(r^k – 1)/(r – 1)] + ar^k by using equation (i) = a[r^k – 1 + r^k(r - 1)]/(r-1) = a[r^k – 1 + r^{k + 1} – r^‑k]/(r - 1) = a[r^{k + 1} – 1]/(r-1) P (n) is true for n = k + 1 Thus, P (n) is true for all n ∈ N.

Prove by the principle of mathematical induction: a + ar + ar2 + … + arn – 1 = a[(rn – 1)/(r – 1)], r ≠ 1

Answer»

Suppose P (n): a + ar + ar² + … + ar^{n – 1} = a[(rⁿ – 1)/(r – 1)]

Now let us check for n = 1,

P (1): a = a (r¹ – 1)/(r - 1)

: a = a

P (n) is true for n = 1.

Then, 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 + … + ar^{k – 1} = a[(r^k – 1)/(r – 1)] … (i)

Therefore,

a + ar + ar² + … + ar^{k – 1} + ar^k

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

= a[(r^k – 1)/(r – 1)] + ar^k by using equation (i)

= a[r^k – 1 + r^k(r - 1)]/(r-1)

= a[r^k – 1 + r^{k + 1} – r^‑k]/(r - 1)

= a[r^{k + 1} – 1]/(r-1)

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

Thus, P (n) is true for all n ∈ N.