Question

Prove by mathematical induction, the following formula for the sum of a geometric series: S(n) r^i = r^(n+1) - 1/r-1 if r is not equal to 1

Answer 1

first do the basis case,

Answer 2

Sum r^i , i=1 .. n = r^(n+1) - 1/r-1

Answer 3

\[\sum_{n}^{i=0} r^i = r ^{n-1} - 1 \div r-1,\] if \[r \neq 1\]

Answer 4

Just to be clear. :-D

Answer 5

did the sum upside down...I'm new here.

Answer 6

How do I show the base case?

Answer 7

ok plug in k=1

Answer 8

also you should probably use parenthesees

Answer 9

So, would the base case be n=0 or n=1?

Answer 10

i believe i=0

Answer 11

Yes, but in order to derive the base case, how would I format it?

Answer 12

so start at i = 0

Answer 13

\[\sum_{i=0}^{n-1} r^i = (r ^{n} - 1)/ (r-1)\ \]

Answer 14

ok. So \[r ^{0}+r^{1} ... r^{n-1} ?\]

Answer 15

the basis case is just one number

Answer 16

lets do the basis case i = 1
then show
r^0 + r^1 = ( r^2 - 1) / ( r - 1 )

Answer 17

actually it might be easier if we do this

Answer 18

\[\sum_{i=0}^{n} r^i = (r ^{n+1} - 1)/ (r-1)\ \]

Answer 19

this is easier to prove, methinks

Answer 20

with me so far?

Answer 21

So n=1, \[r^{0} + r^{1} = (r^{(1+1)} - 1) / (r-1)\]

Answer 22

so let's prove that
r^0 + r^1 + r^2 + ... r^n = ( r^(n+1) - 1 ) ( r-1)

Answer 23

we can use zero
i = 0
r^0 = ( r^(1+0) - 1 ) / ( r - 1 )

Answer 24

but i=1 is fine too, any number in fact

Answer 25

Sorry to be obtuse, but what I'm confused about is that i is already = 0 as stated in the original equation

Answer 26

So are you meaning to change n=0?

Answer 27

'i' is the index, i = 0,1,2,3... n

Answer 28

you dont have to apologize, any questions you can ask. just bear with me, this program is so buggy

Answer 29

Yes. So in this case, it's stated that i=0.

I thought in the base case you're supposed to set n to the lowest possible integer to test teh hypothesis

Answer 30

So if we set n=0 we get:

\[r^{0} = r^{1} - 1 / r-1\]

Answer 31

which is 1=1. Would that be acceptable base case?

Answer 32

yes,

Answer 33

so it passes the basis case.

Answer 34

sorry, the basis test

Answer 35

Does the common ratio (r) not being equal to 1 factor into the basis test at all?

Answer 36

if r = 1 , then we have division by zero

Answer 37

Ok, but the basis test still holds true in this case.

Answer 38

right

Answer 39

because we assumed originally that r != 1

Answer 40

so now to prove by induction, n = k+1 correct?

Answer 41

or rather, we assume that \[\sum_{i=0}^{k} = r^{(k+1)}-1 / r-1\] is true

Answer 42

right, and then show it will be true for
\[\sum_{i=0}^{k+1} = (r^{((k+1)+1)}-1) / r-1\]

Answer 43

and try by induction to prove that \[\sum_{i=0}^{k+1} r^{i} = r^{((k+1)+1)} -1 / r-1\]

Answer 44

exactly

Answer 45

Yayy I'm getting smarter!

Answer 46

you just need parenthesees!

Answer 47

So now, I have to somehow simplify it.

\[\sum_{i=0}^{k+1} r^{i} = r^{0} + r ... r^{k+1} = (r^{k+2} - 1) / (r -1) \]

Answer 48

sometimes its easier to write out the sum
we assume that
1 + r + r^2 + ... + r^k = ( r^(k+1) - 1 ) / ( r - 1)

Answer 49

what if we add r^(k+1) to both sides

Answer 50

Oh wait, yes, I remember my professor doing that.

Answer 51

then the left side is the left hand side of the conclusion

Answer 52

and its a legal algebraic move

Answer 53

Ok, can you explain that one more time? (Thank you so much for your time, this is exactly the clarification i needed)

Answer 54

so i wrote out the assumption (what we assume to be true)
1 + r + r^2 + ... + r^k = ( r^(k+1) - 1 ) / ( r - 1)

Then we stare at the conclusion
1 + r + r^2 + ... +r^k + r^(k+1) = ( r^(k+2) - 1 ) / ( r - 1)

to go from the assumption to the conclusion, it seems natural to add r^k+1 to both sides of the assumption (and thats fine because we assumed it is true, and you can add the same number to both sides of a true equality)

Answer 55

by adding r^k+1 to both sides of the assumption , that will immediately give us the LHS (left hand side) of the conclusion

Answer 56

1 + r + r^2 + ... + r^k = ( r^(k+1) - 1 ) / ( r - 1)
+r^k+1 +r^k+1
--------------------------------------------

Answer 57

OK. so how can we simplify it from there to verify the equality?

Answer 58

ok so we get

Answer 59

1 + r + r^2 + ... + r^k = ( r^(k+1) - 1 ) / ( r - 1)
+r^k+1 +r^k+1
--------------------------------------------
1+ r + r^2 + ... + r^k +1 = ( r^(k+1) - 1 ) / ( r - 1) + r^k+1

Answer 60

now we just have to simplify the right hand side (RHS)

Answer 61

did i lose you?

Answer 62

No, I'm with you so far.

Answer 63

Actually

Answer 64

in the RHS why is the parenthetical \[r^{k+1}\] not \[r^{k+2}\]

Answer 65

because that was what we had in the assumption

Answer 66

but after we simplify we should get r^k+2

Answer 67

Ok so how do we simplify the RHS

Answer 68

because this is where I always screw up

Answer 69

we just need a common denominator