Sequences and series question
(I just need to check the answer)

Question

Sequences and series question 
(I just need to check the answer)

anonymous · Answer

I am getting the answer as  2007006

anonymous · Answer

$$\huge 1^{2}-2^{2}+3^{2}-4^{2}+5^{2}......+2003^{2}$$

anonymous · Answer

what would be the general solution for nth term

anonymous · Answer

@dewfus that would be a very tedious way

anonymous · Answer

how did you solve it

anonymous · Answer

Hint:- Think of a identity

ganeshie8 · Answer

$a^2 - b^2 = (a+b)(a-b)$

anonymous · Answer

What is the answer you are getting 
@ganeshie8

anonymous · Answer

$$\huge -1(3+7+11....2002)+2003^{2}$$
Right?

dan815 · Answer

why not just derive the sum of n^2 formula

anonymous · Answer

It would take much time

ganeshie8 · Answer

anonymous · Answer

Yes so my answer is correct.

anonymous · Answer

there are many ways
like seperating even and odd terms

anonymous · Answer

for instance
1^2 + 3^2 +5^2... 
sum = n/3 (2n+1) (2n-1)

dan815 · Answer

dan815 · Answer

another simple way to subract the series

dan815 · Answer

i think same thing ganeshie did though

ganeshie8 · Answer

ahh reminds me of pascal triangle
$$\large   1^2 + 3^2 +  5^2 + \cdots + (2n-1) =  \binom{2n+1}{3}$$

dan815 · Answer

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