Getting a

this is pair of a^2 - b^2
after breaking it in (a+b)(a-b)

Sum = (1+2) + (3+4) +...(19+20)
Sum = 210

–1^2 + 2^2 – 3^2 + 4^2 – 5^2 + 6^2............-19^2 + 20^2
= (20^2 -19^2) + (18^2 -17^2)+...(2^2 -1^2)
= (20 + 19)(20-19) + (18+17)(18-17)+......+ (2+1)(2-1)
= 20 + 19 + 18 + 17.....+ 3

take the first and last digit,
on applying (a^2-b^2) formula we get
21(19)-21(17)+21(15)-21(13)+...............-21(1)
21(19-17+15-13+11-9+7-5+3-1)
21(2+2+2+2+2)
21(10)=210.

let me try to explain...

work 2 digits by 2 digits

-1^2 +2^2 = 2^2 -1^2 = (2+1)(2-1)

next...

– 3^2 + 4^2 = 4^2 – 3^2 = (4-3)(4+3)

Hence you get (1+2) + (3+4) ....
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Impossible is nothing

OK read it again..and the symbol is just a (-)

so we have -1^2 +2^2 - 3^2+4^2....

can be simplifyed..to

(a-b)(a+b) here b =1 and a=2;

(2-1)(2+1)+(4-3)(4+3)...basically it becomes 1+2+3....+20

so the sum of the first 20 integers is (20+1)/2 * (20-1)+1=210

A it is.

I believe this has already been said in other ways, but I got lost in some of the terminology and symbols, so I had to to break it down a little more for myself. Hopefully this helps anyone still confused by this.

This is a arithmetic sequence, but only if we look at it in pairs. So (-1^2 + 2^2) is the first pair. (-3^2 + 4^2) is the second pair and so on. The sum in each pair adds four to the previous pair each time as follows:

(-1^2 + 2^2) = 3 (note: 3 is our starting number)
(-3^2 + 4^2) = 7
(-5^2 + 6^2) = 11
(-7^2 + 8^2) = 15
.
.
.
(-19^2 + 20^2) = 39

As soon as we recognize the sequence, we can use the sum of n terms arithmetic progression formula:
Sum of n terms = (n/2) x (value of 1st term + value of last term)
substitute:
sum of n terms = (10/2) x (3+39) = 5 x 42 = 210

Note: we used 10 as the N # of terms because we turned the 20 original terms into 10 pairs.

For those concerned about speed, on this problem all that we have to do is calculate the first couple of pairs until we see the pattern, then calculate the last pair (-19^2 + 20^2), add it to the outcome of the first pair (-1^2 + 2^2) and multiply by 5. This can be done in way under 2 minutes.

Hope that helps.

Hi sagarag,

This question has a great "visual component" to it, so I'm going to give you some hints and let you try this question again...

First, let's deal with -1^2 + 2^2

1) 2^2 is the equivalent of a 2x2 square. Draw it and include the 4 individual boxes.
2) -1^2 = -1; Draw a line through one of the 4 squares you just drew. You now have 3 squares left. Notice the pattern in the drawing....

3) Try these same steps again with -3^2 + 4^2; you should end up with a larger drawing but the SAME pattern. How many squares are left here?

4) Can you figure out how many squares would be left with -5^2 + 6^2 WITHOUT drawing the picture this time....? And what about the other 'pairs' up values up to -19^2 + 20^2?

GMAT assassins aren't born, they're made,
Rich
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Hi,

Yes exponents come first...
But -3^2 is not SQUARE of -IVE integer, it is -IVE of square of positive integer..
so you will first square it and then add -ive sign to it..... \(-3^2= -9\)
If it were (-3)^2, this is what would have been the case given by you.. \((-3)^2 = 9\)



If you simplify \(-1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2... +20^2\), you will get 1+2+3+4..+20
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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