Kreshnik Group Title $\\text{How can I find the sum of:} 2001^2-2000^2+1999^2-1998^2+...+3^2-2^2+1^2 ?$ How can I simplify this... ? 2 years ago 2 years ago

1. experimentX

(2001-2000)(2001+2000)+(1999+1998) ... + 3 +2+1

2. experimentX

1+2+3+ .... + 2001 ... i guess so

3. Kreshnik

$2001^2−2000^2+1999^2−1998^2+...+3^2−2^2+1^2?$ this should be better.

4. anonymoustwo44

I got it

5. lgbasallote

it looks like the pattern is x^2 - (x-1)^2 - (x-2)^2 and so on...any ideas on the sum of that experimentx?

6. Kreshnik

(a-b)^2= a^2-2ab+b^2 not a^2-b^2 !!! O_O a^2-b^2=(a-b)(a+b)

7. anonymoustwo44

@Kreshnik fail me :))

8. Kreshnik

@ anonymoustwo44 it's ok ;)

9. Mr.Math

Use this$\sum_{n=1}^k n^2=\frac{1}{6}k(k+1)(2k+1).$

10. experimentX

pattern should be (-1)^(n+1) n^2

11. anonymoustwo44

or you can rewrite that as $\sum_{i=1}^{n}(-1)^{i+1}i^{2}$

12. anonymoustwo44

try it

13. Kreshnik

thank you

14. Rohangrr

mr. 123+1234 is correct

15. lgbasallote

@Mr.Math is that mathematical induction? i would love to learn that..

16. experimentX

(2001-2000)(2001+2000)+(1999+1998)(1999-1998)+ ... + (3+2)(3-2)+1 1(2001+2000)+1(1999+1998)+ ...3+2+1

17. experimentX

1+2+3+ ... + 2001 ... regular ap

18. Mr.Math

Yes, Mathematical induction is the best way to show that the sum above is true. But it's not how it's derived.

19. experimentX

since this is an alternating series, the above formula you mentioned does not work, the method must be similar to AP ... depending on the last term

20. anonymoustwo44

OR since a^2-b^2=(a+b)(a-b) then that could be written as (2001+2000)(2001-2000)+(1999+1998)(1999-1998)+...+(3+2)(3-2)+1 =(2001+2000)(1)+(1999+1998)(1)+...+(3+2)(1)+1 =2001+2000+1999+1998+...+3+2+1 =1+2+3+.....+1998+1999+2000+2001 this is an arithmetic series now we can use the formula $s=(n/2)(a_{1}+a_{n})$ where a sub n is the last term of the series and and a sub 1 is the first term of the series and n is the number of terms in the series so let's compute for the sum... s=(2001/2)(1+2001) s=(2001/2)(2002) s=(2001)(1001) s=2003001 is the sum :D

21. anonymoustwo44

@experiment44 that's just a formula for generating the series above not for solving its sum

22. experimentX

of course ,,,, after you find ap, finding sum is easy as piece of cake

23. experimentX

thanks for medal ... one for you .. for giving complete answer.