1. ## Induction problem

N^2 - (N-1)^2 + (N-2)^2 - (N-3)^2 + ... 1 =
1 + 2 + 3 + ... N

I am trying to prove the statement above by using induction however, I am not sure what the statement means

First I tried to find a base case when N=1

1^2 + 1 = 1 which is not true

2. Originally Posted by hotmail590
N^2 - (N-1)^2 + (N-2)^2 - (n-3)^2 + ... + 1 =
1 + 2 + 3 + ... + n

I am trying to prove the statement above by using induction however, I am not sure what the statement means

First I tried to find a base case when N=1

1^2 + 1 = 1 which is not true

I think you are not providing all the information that you have and we need
to do this. For instance are N and n the same or is one twice the other?

RonL

3. Sorry for the typo, its just one variable N
Also there was not any +'s before 1 and N at the ends

Prove by induction that the sum:

N^2 - (N-1)^2 + (N-2)^2 - (N-3)^2 + ... 1

equals to

1 + 2 + 3 + ... N

4. Originally Posted by hotmail590
Sorry for the typo, its just one variable N
Also there was not any +'s before 1 and N at the ends

Prove by induction that the sum:

N^2 - (N-1)^2 + (N-2)^2 - (N-3)^2 + ... 1

equals to

1 + 2 + 3 + ... N
Base case: the first is 1 when N=1, as is the second, hence the base case
holds.

Now suppose that for some k:

k^2 - (k-1)^2 + (k-2)^2 - (k-3)^2 + ... 1 = k + (k-1) + .. + 1

..............................................=k(k+1)/2

Now:

(k+1)^2 - k^2 + (k-1)^2 - ... 1

...........=(k+1)^2 - (k^2 - (k-1)^2 + (k-2)^2 - (k-3)^2 + ... 1)

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

...........=(k+1)^2 - k(k+1)/2 = k^2/2 + 3k/2 + 1= (k+1)(k+2)/2

...........= (k+1) + k + .. +1.

So if the two expressions are equal for some N=k they are equal for N=k+1,
and they are equal for N=1, hence by induction they are equal for all
N>=1.

RonL

5. Thanks for all your help CaptainBlack. I would just like to make sure that I got the base case for the problem.

So the left side:

N^2 - (N-1)^2 + (N-2)^2 - (N-3)^2 + ... 1

Is equivalent to N(N+1)/2 so when N = 1, (base case) that would equal to 1(1+1)/2 =1

And for the right side:

1+2+3...N

It would be just 1 when N =1 right?

6. Originally Posted by hotmail590
Thanks for all your help CaptainBlack. I would just like to make sure that I got the base case for the problem.

So the left side:

N^2 - (N-1)^2 + (N-2)^2 - (N-3)^2 + ... 1

Is equivalent to N(N+1)/2 so when N = 1, (base case) that would equal to 1(1+1)/2 =1

And for the right side:

1+2+3...N

It would be just 1 when N =1 right?
Not quite, we have

N^2 - (N-1)^2 + (N-2)^2 - (N-3)^2 + ... 1

and when N=1 is just 1, as there is but a single term.

The N(N+1)/2 is present in what I wrote as it is the sum of the first
N integers, that is:

1 + 2 + .. + N = N(N+1)/2,

the proof requires that the sum of the consecutive integers be replaced
by this short from to make the subsequent calculations simpler.

RonL