Number theory - Induction?

Question

Prove by induction that, for a =/ 1 (a NOT EQUAL to 1)

1+a+a^2+...+a^n = (a^n-1)/(a-1)

If someone can please help its much appreciated. Need full working out

Don

Answer 1

Question should read as:-
1 + a + a² ---+ a^n = [ a^(n + 1) - 1] / (a - 1)
Let k be a value of n , thus
P(k):1 + a + a² ---+a^k = [a^(k + 1) - 1] / (a - 1)
Technique is now to show that P(1) is true and P(k + 1) is true.

Consider P(1)
LHS = 1 + a
RHS = (a² - 1) / (a - 1)
RHS = (a - 1) (a + 1) / (a - 1)
RHS = a + 1
Thus P(1) is true.

Consider P(k + 1)
LHS
1 + a + a² ----a^k + a^(k + 1)
RHS
[ a^(k + 1) - 1 ] / (a - 1)
Have now to show that P(k + 1) is true.
Start with P(k) and add same term to each side.
Term to be added is a^(k + 1):-
LHS
1 + a + a² + --a^k + a^(k + 1)
RHS
( a^(k + 1) - 1 ) / (a - 1) + a^(k + 1)
[ a^(k + 1) - 1 + a^(k + 1)(a - 1) ] / (a - 1)
[ a^(k + 1) (1 + (a - 1) ) - 1] / (a - 1)
[ a^(k + 1) (a) - 1 ] / (a - 1)
[ a^(k + 2) - 1 ] / (a - 1)

Thus P(k + 1) is true

P(k) is true
P(1) is true
P(k + 1) is true
THUS P(n) is true.

Answer 2

First, you made a mistake in writing the equation. It should be

1 + a + a^2 + ... + a^n = (a^(n + 1) -1)/(a - 1).

Since "iulydj" did not catch the error, his answer is nonsense.

Here is the proof by induction:

1 + a = (a + 1)*(a - 1)/(a - 1) = (a^2 - 1)/(a - 1), which is the proposition for n = 1.

Now assume that for some k >= 1, we have

1 + a + ... + a^k = (a^(k + 1) - 1)/(a - 1).

That is, we are assuming the theorem is true for n = k. Now consider

1 + a + ... + a^k + a^(k + 1) = (1 + a + ... + a^k) + a^(k + 1)
= (a^(k + 1) - 1)/(a - 1) +a^(k+1)
= [a^(k+1)-1 + (a^(k+1))*(a-1)]/(a - 1)
= [a^(k+1) - 1 + a^(k+2) - a^(k+1)]/(a-1)
= (a^(k + 2) - 1)/(a - 1)
= [a^((k + 1) + 1) - 1]/(a - 1).

That is the theorem for n = k + 1, and that completes the proof.

Answer 3

1. verification:

n = 1 => 1 = (a-1)/(a-1)= 1 TRUE
n = 2 => 1 + a = (a^2 -1)/(a-1)= (a-1)(a+1)/(a-1) = a + 1 True:
2. suppose : 1+a+a^2+...+a^n = (a^n-1)/(a-1)
3. demonstrate : 1+a+a^2+...+a^n + a^(n+1)= (a^(n+1)-1)/(a-1)

1+a+a^2+...+a^n + a^(n+1)= (a^(n+1)-1)/(a-1)
1+a+a^2+...+a^n = (a^n-1)/(a-1)

=> (a^n-1)/(a-1) + a^(n+1)= (a^(n+1)-1)/(a-1)
(a^n-1)/(a-1) + a^(n+1)*(a-1)/(a-1)= (a^(n+1)-1)/(a-1) |: (a-1)
a^n-1+ a^(n+1)*(a-1) = a^(n+1)-1
a^n+ a^(n+1)*(a-1) = a^(n+1)
a^n+ a^(n+2) - a^(n+1) = a^(n+1)
a^n+ a^(n+2) - 2a^(n+1) = 0
a^n(1 + a^2 - 2a)= 0
a^n( a^2 - 2a + 1)= 0
a^n(a-1)(a-1) = 0 ....

Answer 4

I think that this should be
1+a+a^2+...+a^n=(a^(n+1)-1)/(a-1)
For n= 0:
1 ? (a^(0+1)-1)/(a-1)
1 ? (a-1)/(a-1)
1 ? 1 These are equal, so it is true for n=1
Assume it is true for n:
1+a+a^2+...+a^n=(a^(n+1)-1)/(a-1)
Now add a^(n+1) to both sides
1+a+a^2+...+a^n+a^(n+1)=(a^(n+1)-1)/(a-1)+a^(n+1)
=(a^(n+1)-1)/(a-1)+(a^(n+1))*(a-1)/(a-1)
=(a^(n+1)-1)/(a-1)+(a^(n+2)-a^(n+1))/(a-1)
=(a^(n+1)-1+a^(n+2)-a^(n+1))/(a-1)
=(a^(n+2)-1)/(a-1)
=(a^(n+1+1)-1)/(a-1)
Which is exactly the formula for (n+1)

Answer 5

I'm sure that an induction proof would be valuable, but this looks to me like a simple geometric series.
S=a1(1-r^q)/(1-r)
a1=1 r=a q=# of terms=n+1
S=1(1-a^(n+1))/(1-a)
S=(a^(n+1)-1)/(a-1)