Question

Math Brain Twister:

100^2 - 99^2 + 98^2 - 97^2 + 96^2 - 95^2 + ... + 2^2 - 1^2 = ?

Anyone?
I'll give medal.. :))

Answer 1

its like
100+99+98+97+....+2+1

Answer 2

n(n+1)/2

Answer 3

sum of special series?

Answer 4

n= 100

Answer 5

5050

Answer 6

no

Answer 7

100^2 - 99^2 = (100+99)(100-99) = 100+99

Answer 8

hartnn is crrect @liliegirl

Answer 9

Solution
Since you can notice that there are fifty pairs of n^2 – (n-1) ^2,
n^2 – (n-1)^2 = n + (n – 1)

Thus 100^2 - 99^2 + 98^2 - 97^2 + 96^2 - 95^2 + ... + 2^2 - 1^2 can also be written as
100 + 99 + 98+ ... + 2 + 1 = (100 x 101)/2 = 5050

Answer 10

Thanks! @hartnn

Answer 11

welcome ^_^

Answer 12

You guys are just using memory. Your explanations don't make sense to anyone who isn't familiar with the problem already.\[
n^2-(n-1)^2=n^2-(n^2-2n+1) = 2n-1
\]

Answer 13

So if you're summing over \(2n-1\) when stepping by \(2\), how does that imply you are summing over \(n\) when stepping by \(1\)?