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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.

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.

I dont see how you've applied the (a^2-b^2) formula?

Thanks for reposting the question. I couldnt make sense of the symbol either. I am new to GMATClub and I must say that this group is more than just impressive. I appreciate the camaraderie the members show to help each other get their best, and hope to contribute whenever I can. In that spirit....

I got the (2-1)(2+1) + (4-3)(4+3) + (6-5)(6+5) + (8-7)(8+7) + ..... + (20-19)(20+19).... part But then this translates to 3 + 7 + 11 + 15 +...... + 39, which I solved as 10 terms with next term incremented by 4. So their sum would be the middle term (average of 19 and 23 = 21) multiplied by 10 = 210.

Using "Sum of n terms of an equally spaced series = middle terms * n" (if n = even, middle term is average of the two middle terms).

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:

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.

if you didn't see that grouping you could always after doing a couple of numbers see that after every addition step which happens to coincide with even numbers there is an overall increase but the value never attains the value of the squared term added...

so you know that after adding 20^2 the answer will be less than 400

at this point, either guess or do a few terms and see that the negative value build up to higher than 70.....ie 400-330.....

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?

Find the the sum of the first 20 terms of this series which [#permalink]

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02 Apr 2016, 18:58

i got to A..but my main concern, does PEMDAS apply here at all or not? if not, it needs to be specified so. PEMDAS - parenthesis, exponents, multiplication, division, addition, subtraction. exponents come first, so any NEGATIVE integer squared is a positive number...no? my approach, though lengthy: list all squares (good to remember squares of the first 20 integers): 1 is positive 4 positive 9 negative 16 positive 25negative 36positive 49negative 64positive 81negative 100positive 121negative 144positive 169negative 196positive 225negative 256positive 289negative 324positive 361negative 400positive

now 400-361 = 39 324-289= 35 256-225=31 196-169=27 144-121=23 100-81=19 64-49=15 36-25=9 16-9=7 and +14=5 now 39+35+31+27+23+19+15+9+7+5 group to be easier: 35+15=50 27+23=50 31+39=70 19+9=28 7+5=12

last digit is 0, so C is out 50+50+70+40=210 A

p.s. surprisingly or not, the sum of all numbers between 1 and 20, inclusive is 20(21)/2 = 210.

gmatclubot

Find the the sum of the first 20 terms of this series which
[#permalink]
02 Apr 2016, 18:58

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