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# If the sum of the first n positive odd integers is n^2, what is the su

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Intern
Joined: 02 Dec 2015
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If the sum of the first n positive odd integers is n^2, what is the su  [#permalink]

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Updated on: 21 Feb 2016, 09:28
2
15
00:00

Difficulty:

65% (hard)

Question Stats:

61% (02:06) correct 39% (02:31) wrong based on 161 sessions

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If the sum of the first n positive odd integers is n^2, what is the sum of all odd numbers from 25 to 79?

A) 39^2-11^2
B) 39^2-12^2
C) 40^2-12^2
D) 79^2-22^2
E) 40^2-11^2

The answer is C but, I don't know how..

Originally posted by maverickjin8 on 02 Dec 2015, 05:27.
Last edited by chetan2u on 21 Feb 2016, 09:28, edited 2 times in total.
Formatted the question, added the OA and renamed the topic.
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Re: If the sum of the first n positive odd integers is n^2, what is the su  [#permalink]

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02 Dec 2015, 06:01
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maverickjin8 wrote:
If the sum of the first n positive odd integers is n^2, what is the sum of all odd numbers from 25 to 79?

A) 39^2-11^2
B) 39^2-12^2
C) 40^2-12^2
D) 79^2-22^2

The answer is C but, I don't know how..

Follow posting guidelines (link in my signatures). Any and all analyses should either be put under spoilers or in the next post.

As for this question, look below.

You can solve the question in 2 ways:

Method 1: Sum of an arithmetic progression (=a series such as a1 a2 a3 a4 .... such that a1-a2 = a2-a3 = a3-a4 = constant, difference between consecutive terms remains constant.) =

S = $$\frac {n[2a+(n-1)*d]}{2}$$, where n = number of terms, a = 1st term, d = constant difference

For the question asked, odd number from 25 to 79 are 25,27,29,31...77,79 and any 2 consecutive terms differ by 2. Thus, n = (79-25)/2+1 = 28, d = 2, a = 25.

Therefore, Sum = S = $$\frac {n[2a+(n-1)*d]}{2}$$ = $$\frac {28[2*25+(28-1)*2]}{2}$$ = $$28*52$$= $$(40-12)*(40+12)$$ = $$40^2 - 12^2$$ (as $$(a+b)*(a-b) = a^2-b^2$$).

Alternately, Method 2, you are given that sum of n odd integers = $$n^2$$.

Thus Sum of odd integers from 25 to 79 = Sum of odd integers from 1 to 79 - Sum of odd integers from 1 to 23 (chose 23 as we need 25 in our final calculations).

==> S = $${n_1}^2 - {n_2}^2$$, where $$n_1$$ = 40 and $$n_2$$= 12 ==> S = $$40^2 - 12 ^2$$.

Thus C is the correct answer.
##### General Discussion
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Re: If the sum of the first n positive odd integers is n^2, what is the su  [#permalink]

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02 Dec 2015, 06:27
It was so simple... I can't believe I've didn't even tried to solve it. Thank you very much.
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Re: If the sum of the first n positive odd integers is n^2, what is the su  [#permalink]

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19 Feb 2016, 20:36
oh man..spent some additional seconds because I did not look at the answer choices..
79 is the 40th odd number
to find the sum of the odd numbers from 25 to 79, we need to deduct from the sum of the first 40 odd numbers, the sum of the first 12 odd numbers. note that 25 is the 13th odd number, thus, 23 is the 12.

40^2 - 12^2
i went even further, and expanded it as: (40+12)(40-12)=52x28...
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Re: If the sum of the first n positive odd integers is n^2, what is the su  [#permalink]

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18 Mar 2017, 13:35
79-25=54/2=27+1 - 28 odd numbers
28*104/2=28*52 (40-12)*(40+12)=40^2-12^2
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If the sum of the first n positive odd integers is n^2, what is the su  [#permalink]

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10 Mar 2019, 06:53
maverickjin8 wrote:
If the sum of the first n positive odd integers is n^2, what is the sum of all odd numbers from 25 to 79?

A) 39^2-11^2
B) 39^2-12^2
C) 40^2-12^2
D) 79^2-22^2
E) 40^2-11^2
[/spoiler]

I did it like this:

1) 79 - 25 = 54 --> 54/2 = 27 --> + 1 --> 28 odd ints (since first and last numbers are odd, there's 1 more odd than even in this set of consecutive positive ints)
2) Sum of First+Last ints is 25+79 = 104. There are 28 odd ints, so 28/2 = 14 such pairs --> 14*104 = 1456.
3) Target is 1456. Check the answers, since our number is Even we can immediately eliminate B, D, E (can only be E-E or O-O to get an Even answer)
4) C jumps out as being right, since 40^2 is 1600 which is close to the target. Check this one: 1600 - 144 = 1456. Correct.

It would have been good for me to immediately realize that Sum n^2 is 79/2 --> 39R1 --> 40^2
Then 25/2 = 12R1 - 1 = 12^2
i.e. (Large number/2 + 1)^2 - (Small number/2 - 1)^2 (because otherwise we'll lose the term 25 when we subtract, since "all odd numbers" means inclusive).
If the sum of the first n positive odd integers is n^2, what is the su   [#permalink] 10 Mar 2019, 06:53
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