$$\text{How can I find the sum of:} 2001^2-2000^2+1999^2-1998^2+...+3^2-2^2+1^2 ?$$

How can I simplify this... ?

Question

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

experimentx · Answer

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

experimentx · Answer

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

anonymous · Answer

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

anonymous · Answer

I got it

lgbasallote · Answer

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?

anonymous · Answer

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

anonymous · Answer

@Kreshnik fail me :))

anonymous · Answer

@ anonymoustwo44 it's ok ;)

mr.math · Answer

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

experimentx · Answer

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

anonymous · Answer

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

anonymous · Answer

try it

anonymous · Answer

thank you

anonymous · Answer

mr. 123+1234 is correct

lgbasallote · Answer

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

experimentx · Answer

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

experimentx · Answer

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

mr.math · Answer

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

experimentx · Answer

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

anonymous · Answer

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

anonymous · Answer

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

experimentx · Answer

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

experimentx · Answer

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