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Find the the sum of the first 20 terms of this series which
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Updated on: 15 Jun 2010, 11:33
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Find the the sum of the first 20 terms of this series which begins this way : 1^2 + 2^2  3^2 + 4^2  5^2 + 6^2... (a) 210 (b) 330 (c) 519 (d) 720 (e) 190 Show working please!
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Originally posted by MBAlad on 02 Dec 2006, 19:03.
Last edited by alexsr on 15 Jun 2010, 11:33, edited 2 times in total.



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Getting a
this is pair of a^2  b^2
after breaking it in (a+b)(ab)
Sum = (1+2) + (3+4) +...(19+20)
Sum = 210



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I'm missing something here 1^2 = 1, 2^2 = 4 3^2 = 9
So I'm getting 1 + 4 + 9...........?????



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â€“1^2 + 2^2 â€“ 3^2 + 4^2 â€“ 5^2 + 6^2............19^2 + 20^2
= (20^2 19^2) + (18^2 17^2)+...(2^2 1^2)
= (20 + 19)(2019) + (18+17)(1817)+......+ (2+1)(21)
= 20 + 19 + 18 + 17.....+ 3



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anindyat wrote: â€“1^2 + 2^2 â€“ 3^2 + 4^2 â€“ 5^2 + 6^2............19^2 + 20^2 = (20^2 19^2) + (18^2 17^2)+...(2^2 1^2) = (20 + 19)(2019) + (18+17)(1817)+......+ (2+1)(21) = 20 + 19 + 18 + 17.....+ 3
I'm half way there:
How does (20+19)(2019) + (18+17)(1817)+...... turn into 20+19+18.....??
How did you arrive at 210
Urggghhh I'm having a bad day!



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take the first and last digit,
on applying (a^2b^2) formula we get
21(19)21(17)+21(15)21(13)+...............21(1)
21(1917+1513+119+75+31)
21(2+2+2+2+2)
21(10)=210.



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Hi pzazz
Quote: take the first and last digit, on applying (a^2b^2) formula we get 21(19)21(17)+21(15)21(13)+...............21(1) 21(1917+1513+119+75+31) 21(2+2+2+2+2) 21(10)=210.
I dont see how you've applied the (a^2b^2) formula?
21(19) = 399?
Also why did you use 21?????
I will get there!!!!



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let me try to explain...
work 2 digits by 2 digits
1^2 +2^2 = 2^2 1^2 = (2+1)(21)
next...
â€“ 3^2 + 4^2 = 4^2 â€“ 3^2 = (43)(4+3)
Hence you get (1+2) + (3+4) ....
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Sorry for reviving an old thread, but isn't there an easier way to do this since it's related to arithmetic series?
We know that 2^21^2 = 3
4^2  3^2 = 7
6^2  5^2 = 11
and so forth.
So you have a series of 10 numbers, starting with a1 = 3, d = 4, and n = 10.
Sum of this series is [10( 2(3) + (101)(4)]/2 = 210.



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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
(ab)(a+b) here b =1 and a=2;
(21)(2+1)+(43)(4+3)...basically it becomes 1+2+3....+20
so the sum of the first 20 integers is (20+1)/2 * (201)+1=210
A it is.



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Re: Yet Another Sequence!
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29 Jun 2009, 13:13
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 (21)(2+1) + (43)(4+3) + (65)(6+5) + (87)(8+7) + ..... + (2019)(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).



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Re: Yet Another Sequence!
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14 Apr 2010, 10:17
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.



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Re: Yet Another Sequence!
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09 Jul 2010, 20:26
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 400330.....



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Re: Find the the sum of the first 20 terms of this series which
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06 Jun 2015, 15:49
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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Find the the sum of the first 20 terms of this series which
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02 Apr 2016, 17:58
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 400361 = 39 324289= 35 256225=31 196169=27 144121=23 10081=19 6449=15 3625=9 169=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.



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Re: Find the the sum of the first 20 terms of this series which
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03 Apr 2016, 02:08
mvictor wrote: 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 400361 = 39 324289= 35 256225=31 196169=27 144121=23 10081=19 6449=15 3625=9 169=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. Hi, Quote: exponents come first, so any NEGATIVE integer squared is a positive number...no? 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.. \((3)^2 = 9\)Quote: p.s. surprisingly or not, the sum of all numbers between 1 and 20, inclusive is 20(21)/2 = 210. 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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Re: Find the the sum of the first 20 terms of this series which
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12 Aug 2016, 08:14



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Re: Find the the sum of the first 20 terms of this series which
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21 Sep 2018, 14:26
MBAlad wrote: Find the the sum of the first 20 terms of this series which begins this way : 1^2 + 2^2  3^2 + 4^2  5^2 + 6^2...
(a) 210 (b) 330 (c) 519 (d) 720 (e) 190
Show working please! Arithmetic progression )



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Find the the sum of the first 20 terms of this series which
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21 Sep 2018, 19:46
MBAlad wrote: Find the the sum of the first 20 terms of this series which begins this way : 1^2 + 2^2  3^2 + 4^2  5^2 + 6^2...
(a) 210 (b) 330 (c) 519 (d) 720 (e) 190
Show working please! if t1+t2=3, t3+t4=7, and t5+t6=11, then we can assume a difference of 4 between consecutive term pairs thus, t9+t10=19 19+4/2=21=median of series 10 term pairs*21 median=210 A




Find the the sum of the first 20 terms of this series which &nbs
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