## 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 Group Title

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

2. experimentX Group Title

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

3. Kreshnik Group Title

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

4. anonymoustwo44 Group Title

I got it

5. lgbasallote Group Title

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 Group Title

(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 Group Title

@Kreshnik fail me :))

8. Kreshnik Group Title

@ anonymoustwo44 it's ok ;)

9. Mr.Math Group Title

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

10. experimentX Group Title

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

11. anonymoustwo44 Group Title

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

12. anonymoustwo44 Group Title

try it

13. Kreshnik Group Title

thank you

14. Rohangrr Group Title

mr. 123+1234 is correct

15. lgbasallote Group Title

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

16. experimentX Group Title

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

17. experimentX Group Title

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

18. Mr.Math Group Title

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 Group Title

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 Group Title

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 Group Title

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

22. experimentX Group Title

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

23. experimentX Group Title

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