a+ ar + ar2 + + arn-1= a(rn-1)/r-1 prove this by - Maths - Principle of Mathematical Induction

Question

a+ ar + ar2 + + arn-1= a(rn-1)/r-1
prove this by PMI

Nikhil Kumar Singh · Accepted Answer

Dear Student,

We need to prove that

$S_{n} = a + a r + a r^{2} + ... + a r^{n - 1} = \frac{a (1 - r^{n})}{1 - r}, r \neq 1, n \in N$ ........(1)

using principle of Mathematical induction.

Let n = 1.

Plug-in this value in equation (1).

Thus,

$S_{1} = \frac{a (1 - r)}{1 - r} = a$

Thus, equation (1) is true for n = 1.

Let $S_{m}$ is true.

Then,

$S_{m} = a + a r + a r^{2} + ... + a r^{m - 1} = \frac{a (1 - r^{m})}{1 - r}, r \neq 1, m \in N$ is true.

We need to prove that $S_{m + 1}$ is true.

That is we need to prove that,

$S_{m + 1} = a + a r + a r^{2} + ... + a r^{m} = \frac{a (1 - r^{m + 1})}{1 - r}, r \neq 1, m + 1 \in N$ is true.

Consider the L.H.S of the above equation.

$S_{m + 1} = a + a r + a r^{2} + ... + a r^{m - 1} + a r^{m}$

Since the statement $S_{m} = a + a r + a r^{2} + ... + a r^{m - 1} = \frac{a (1 - r^{m})}{1 - r}, r \neq 1, m \in N$ is true for $n = m$ , we can rewrite $S_{m + 1}$ :

Thus,

$S_{m + 1} = S_{m} + a r^{m} = \frac{a (1 - r^{m})}{1 - r} + a r^{m} = \frac{a (1 - r^{m}) + a r^{m} (1 - r)}{1 - r} = \frac{a - {a r}^{m} + a r^{m} - a r^{m + 1}}{1 - r} = \frac{a - a r^{m + 1}}{1 - r} = \frac{a (1 - r^{m + 1})}{1 - r}$

Thus, if $S_{m}$ is true then $S_{m + 1}$ is true.

Thus, $S_{n} = a + a r + a r^{2} + ... + a r^{n - 1} = \frac{a (1 - r^{n})}{1 - r}, r \neq 1, n \in N$ is true for all $n \in N$ .

Regards

a+ ar + ar2 + ...... + arn-1= a(rn-1)/r-1 prove this by PMI .

a+ ar + ar² + ...... + ar^n-1= a(rⁿ-1)/r-1 prove this by PMI .