0
$\begingroup$

Determine the value of: $$(1^2+3^2+5^2+...+99^2)-(2^2+4^2+6^2+...+100^2)+(4+8+12+...+200)$$

Attempt

I can see that the first 2 series are neither arithmetic nor geometric, however when I used the equation for an arithmetic series, I got the right answer (50).

My teacher told the class a hint: look at the first term of each bracket together, then the second term of each bracket, and so on. I don’t understand why we need to do this and although I know that the way I solved the problem was incorrect, I can’t think of any other way to solve it.

$\endgroup$
3
  • $\begingroup$ What is the pattern for $2^2 +4^2 + 6 +2 + .... + 100$? Was that supposed to be $2^2 + 4^2 + 6^2 + .... + 100$? $\endgroup$ – fleablood Feb 14 '19 at 20:08
  • $\begingroup$ "however when I used the equation for an arithmetic series, I got the right answer " What did you do about the first two series? Ignore them? " I know that the way I solved the problem was incorrect" It'd help if you told us how you solved it. We have no idea what you did, so we have no idea if it was correct or not. $\endgroup$ – fleablood Feb 14 '19 at 20:26
  • $\begingroup$ " I don’t understand why we need to do this " Well, if you tried it you would have seen. $(1^2 - 2^2 + 4) + (3^2 - 4^2 + 8) + (5^2 - 6^2 + 12)+ ....$ what is $1^2 -2^2 + 4$? What is $3^2 -4^2 + 8$? Etc. Can you see why that might be? $\endgroup$ – fleablood Feb 14 '19 at 20:30
4
$\begingroup$

$$1^2 -2^2 +4= (1-2)(1+2) +4 = -3 +4 =1 \\ 3^2 - 4^2 + 8 = (3-4)(3+4)+8 = -7 + 8 =1$$ See the pattern?

The key to this (and a lot of problems like this) is $n^2 - m^2 = (n-m)(n+m)$. It's also just a really helpful identity in general.

$\endgroup$
2
$\begingroup$

So I think the hint is to rearrange the expression as

$$ (1^2-2^2)+(3^2-4^2)+\dots+(99^2-100^2) + 4(1 + 2 + \dots + 50) $$

Does that help?

$\endgroup$
1
$\begingroup$

$(1^2 + 3^2 + ... + 99^2) = \sum_{k=1}^{50} (2k-1)^2$

$(2^2 + 4^2 + .... + 100^2) = \sum_{k=1}^{50} (2k)^2$

$(4 + 8 + ..... + 200) = \sum_{k=1}^{50} 4*k$

So $(1^2 + 3^2 + ... + 99^2)-(2^2 + 4^2 + .... + 100^2)+(4 + 8 + ..... + 200) =$

$\sum_{k=1}^{50}[ (2k-1)^2 - (2k)^2+ 4*k] =$

$\sum_{k=1}^{50}[( 4k^2 - 4k + 1) -4k^2 + 4*k] =$

$\sum_{k=1}^{50}1 = 50$.

That's it!

Let try a few.

$1^2 - 2^2 + 4 = 1$

And $3^2 - 4^2 + 8 = 1$

And $5^2 - 6^2 + 12 = 1$ and ... interesting.

But $7^2 - 8^2 + 16 = (7-8)(7+8) + 16 = -15 + 16 = 1$ or

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

Lot's of ways to do it but $(2k-1)^2 + (2k)^2 + 4k = 1$. Always.

$\endgroup$
1
  • $\begingroup$ Thank you to everyone who replied, it was very helpful! :) $\endgroup$ – Sophie04 Feb 14 '19 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.