Help with proof by induction problem

Question

Any input on this problem I'm working on?

I want to prove by induction that for all natural $n$ we have that

$\sum_{i=1}^n -1^jj^2 = (-1)^n\frac{n(n+1)}{2}$

So far I have:

1) Proved the base case, n = 1, which gives me:

$\sum_{i=1}^n -1^jj^2 = (-1)\frac{2}{2}$

$\rightarrow\sum_{i=1}^n -1^jj^2 = -1$

Which makes sense.

2) For the induction step, I assume n = k is true, I get:

$\sum_{i=1}^n -1^jj^2 = (-1)^k\frac{k(k+1)}{2}$

Then I want to find out if this is true for n = k+1. I insert for k+1.

$\sum_{i=1}^n -1^jj^2 = (-1)^k+1\frac{k+1((k+1)+1)}{2}$

I know you guys usually ask for me making an effort at it, but I am literally stuck, and have been so for some time. Any tips?

BTW I have read this guide, which was a good read, but didn't really help me figuring out how to solve these. Please share other links if you have any.

Please post your problem here with appropriate formatting in MathJax (tutorial: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) instead of linking to an image of the problem. — Alex Ortiz, Aug 29 '16 at 18:38
Possible duplicate of Sum of alternating sign squares of integers stuck with proof by induction — JMoravitz, Aug 29 '16 at 18:48

score 1 · Answer 1 · answered Aug 29 '16 at 18:50

Hint:

$$\sum_{j=1}^{k+1}(-1)^j j^2 = \left[\sum_{j=1}^{k}(-1)^j j^2\right] + (-1)^{k+1}(k+1)^2.$$

Now you can use the assumption on the bracketed term.

You may need to consider even and odd $k$ separately.

(Note: I'm not sure $-1^k$ and $(-1)^k$ are the same thing. The second one makes it clear what you mean.)

score 0 · Answer 2 · answered Aug 29 '16 at 18:50

Here you are,

Assume: $\sum_{i=1}^k -1^jj^2 = (-1)^k\frac{k(k+1)}{2}$

Using the hypothesis show that:

$\sum_{i=1}^{k+1} -1^jj^2 = (-1)^{k+1}\frac{k+1((k+1)+1)}{2}$

the part you are missing:

$\sum_{i=1}^{k+1} -1^jj^2 = (\sum_{i=1}^{k} -1^jj^2) + (-1)^{k+1}(k+1)^2$

Now substitute based on the inductive hypothesis. Apply your algebra skills and take it home.

score 0 · Answer 3 · answered Aug 29 '16 at 18:51

0

$$\sum_{j=1}^{k+1}(-1)^j j^2=(-1)^{k+1}(k+1)^2+\sum_{j=1}^{k}(-1)^j j^2=(-1)^{k+1}(k+1)^2+(-1)^k\frac{k(k+1)}{2}$$ therefore $$\sum_{j=1}^{k+1}(-1)^j j^2=(-1)^{k+1}\left((k+1)^2-\frac{k(k+1)}{2}\right)$$ $$\qquad=(-1)^{k+1}\frac{(k+1)(k+2)}{2}$$

answered Aug 29 '16 at 18:51

Behrouz Maleki

11,182

score 0 · Answer 4 · answered Aug 29 '16 at 18:51

When dealing with induction proofs, the usual strategy is to try to find some way to reduce the expression we get when inserting $n=k+1$ to the expression when $n=k$ so we can apply the assertion given by the inductive step and then work from there.

If the expression in question is a sum (as it is the case) it is straightforward to to just that, as we can pop the last term of the summation to get the expression when $n=k$.

Thus: $$ \sum_{i=1}^{k+1} (-1)^jj^2 = (-1)^{k+1}(k+1)^2 + \sum_{i=1}^{k} -1^jj^2= (-1)^{k+1}(k+1)^2 + (-1)^k\frac{k(k+1)}{2} $$

Now it is a matter of applying algebra.

Help with proof by induction problem

4 Answers4