Proving 12 - 22 + 32 - 42 + ... + (-1)n+1)n2 for Every Positive Integer n

[magic88]

Homework Statement

This is a proof problem in the mathematical induction section of my textbook. I am having trouble with question (c).

Result: 1² + 2² + 3² + ... + n² = n(n+1)(2n+1)/6 for every positive integer n.

(a) Use the result to determine the formula for 2² + 4² + 6² + ... + (2n)² for every positive integer n.

• I have already found the solution to this problem: 2² + 4² + 6² + ... + (2n)² = (2/3)n(n+1)(2n+1) for every positive integer n.

(b) Use the result to determine the formula for 1² + 3² + 5² + ... + (2n -1)² for every positive integer n.

• I have already found the solution to this problem: 1² + 3² + 5² + ... + (2n -1)² = (2/3)n(n+1)(2n+1) - 2n(n+1) + n for every positive integer n.

(c) Use (a) and (b) to determine the formula for 1² - 2² + 3² - 4² + ... + (-1)ⁿ⁺¹)n² for every positive integer n. Use mathematical induction to prove this.

Homework Equations

Principle of Mathematical Induction: For each positive integer n, let P(n)be a statement. If P(1) is true and the implication "If P(k), then P(k+1) is true," then P(n) is true for every positive integer n.

The Attempt at a Solution

I have found the solutions to part (a) and (b) and have proved them. (Lots of algebra involved!) But I've been trying to figure out part (c) for about an hour now, and I just don't know how to approach it.

I started by listing out a few of the sums from part (c):

• When n = 1, sum = 1
• When n = 2, sum = -3
• When n = 3, sum = 6
• When n = 4, sum = -10
• When n = 5, sum = 15
• When n = 6, sum = -21

And listing out sums from part (a):

• When n = 1, sum = 4
• When n = 2, sum = 20
• When n = 3, sum = 56
• When n = 4, sum = 120

And listing out sums from part (b):

• When n = 1, sum = 1
• When n = 2, sum = 10
• When n = 3, sum = 35
• When n = 4, sum = 84

I found a pattern between the three sums:

• Part (c) sum = -3 = 1-4
• Part (c) sum = 6 = 10-4
• Part (c) sum = -10 = 10-20
• Part (c) sum = 15 = 35-20
• Part (c) sum = -21 = 56-35

(I’m sorry, I don’t really know how to explain the pattern, but if you look back at the lists of sums I made above, I think you will understand what I mean.)

I don’t know how to proceed from here. Any help is GREATLY appreciated! Please let me know if you need any clarification.

Kendra :)

 

[Mmmm]

(c)=(b)-(a) with 2n replaced with n.

also, for your previous proofs for which you said you used lots of algebra, consider this:
you have

\sum_{r=1}^{n}r^{2}=\frac{1}{6}n(n+1)(2n+1)\]

(a) find

\sum_{r=1}^{n}(2r)^{2}\]
(write this in terms of the first sum)

(b) find

\sum_{r=1}^{n}(2r+1)\]
(write this in terms of the first sum too)

That should make things considerably easier

 

[magic88]

Thanks very much! Your tip about writing (a) and (b) in terms of the first sum really helped... I don't know why I didn't think of that.

I got:

(-1)^{n+1}\frac{n^2+n}{2}

 

[Mmmm]

Thanks very much! Your tip about writing (a) and (b) in terms of the first sum really helped... I don't know why I didn't think of that.

I got:

(-1)^{n+1}\frac{n^2+n}{2}

No problem, glad it helped... Now all you have to do is prove your answer for (c) using induction.