Question

1+r+r²+...........+r^n+1=1-r^n/1-r​mathematical induction mesthod

Answer 1

Answer + Step-by-step explanation:

the correct question:

For r ≠ 1 ,Prove using the mathematical induction method that :

$1+\cdots+r^n =\frac{1-r^{n+1}}{1-r}$

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for n = 0 :

1⁰ = 1  and  (1 - r⁰⁺¹)/(1 - r) = (1 - r)/(1 - r) = 1

Then the property is true for n = 0.

For n ≥ 0 :

Suppose

$1+\cdots+r^n =\frac{1-r^{n+1}}{1-r}$

And prove that

$1+\cdots+r^{n+1} =\frac{1-r^{n+2}}{1-r}$

Since :

$1+\cdots+r^{n+1} =(1+\cdots+r^n)+r^{n+1}$

Then

$1+\cdots+r^n+r^{n+1} =\frac{1-r^{n+1}}{1-r}+r^{n+1}$

$= \frac{1-r^{n+1}+r^{n+1}(1-r)}{1-r}$

$= \frac{1-r^{n+2}}{1-r}$

Then according to the mathematical induction method

$1+\cdots+r^n =\frac{1-r^{n+1}}{1-r}$

Where n is a natural number and r ≠ 1.