What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? : Problem Solving (PS)

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

18 May 2020, 05:21

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Bunuel wrote:

What is the value of \(1^2 + 3^2 + 5^2 + ... +\) \(3^{12}\)?

A. 9,455
B. 5,456
C. 4,892
D. 4,792
E. 3,468

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hi, Bunuel
The highlighted part shouldn't be \(31^{2}\) ?

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

18 May 2020, 05:26

Expert Reply

lacktutor wrote:

Bunuel wrote:

What is the value of \(1^2 + 3^2 + 5^2 + ... +\) \(3^{12}\)?

A. 9,455
B. 5,456
C. 4,892
D. 4,792
E. 3,468

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hi, Bunuel
The highlighted part shouldn't be \(31^{2}\) ?

_______________________
Yes. Edited. Thank you.

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What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

18 May 2020, 05:39

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\(N= [1^2+2^2+3^2+4^2+........30^2+31^2]- [2^2+4^2+6^2+......+30^2]\)

\(N= [1^2+2^2+3^2+4^2+........30^2+31^2]- 2^2 [1^2+2^2+3^2+......+15^2]\)

Sum of squares of n consecutive terms \(= \frac{n(n+1)(2n+1)}{6}\)

\(N = \frac{31*32*63}{6} - 4*\frac{(15)(16)(31)}{6}\)

N = 31*16*21 - 10*16*31

N= 31*16(21-10) = 31*16*11

Unit digit of product is 6

B

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What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

18 May 2020, 06:03

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What is the value of \(1^{2}+3^{2}+5^{2}+...+31^{2} \)?

(1+ 9+ 25+ 49+ 81 )+ (121+ 169+ 225+ 289+ 361 ) + (441+ 529+ 625+ 729+ 841) + 961 = ???

Sum up the units digits:
---> \(..25+ 25 + 25 +1 = ...6\)

Answer (B).

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

19 May 2020, 00:36

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See the attachment. Can somebody please tell me how to solve this problem with whiteboard in under 2 minutes.

Attachments

1.PNG [ 23.77 KiB | Viewed 7142 times ]

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

19 May 2020, 01:04

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Bunuel wrote:

What is the value of \(1^2 + 3^2 + 5^2 + ... + 31^2\)?

A. 9,455
B. 5,456
C. 4,892
D. 4,792
E. 3,468

Of course, there is a formula for this,
Sum of square of odd numbers \(1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3}\)
Here 2n-1=31....n=16
So \(SUM = \frac{n(2n-1)(2n+1)}{3}=\frac{16*31*33}{3}=16*31*11\)=5456

Also we can almost always do such question by units digit or some other pattern.

\(1^2 + 3^2 + 5^2 + ... + 31^2\) The units digit here is a set of (1^2+3^2+5^2+7^2+9^2), which will have same units digit as 1+9+5+9+1=15 or 5

Now till 30^2, we have 3 such sets, so units digit of \(1^2 + 3^2 + 5^2 + ... + 30^2\) will be 5, and when we add 31^2 to it, the units digit becomes 5+1=6

Only B possible.

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

19 May 2020, 01:43

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sambitspm wrote:

See the attachment. Can somebody please tell me how to solve this problem with whiteboard in under 2 minutes.

sambitspm

Look at unit digit of the results

\(1^2 = 1\)
\(3^2 = 9\)
\(5^2 = 5\)
\(7^2 = 9\)
\(9^2 = 1\)
\(11^2 = 1\)
\(13^2 = 9\)
\(15^2 = 5\)

i.e. Unit digits are cyclic

Now, we have 16 odd squares and each cycle has 5 numbers with unit digit sum = 1+9+5+9+1 = 5

So 16 terms will have unit digit sum = 5+5+5+1(for 16th terms) = 6

Answer: Option B

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

19 May 2020, 01:58

Thanks GMATinsight. This is good trick, but what if we had two options ending in-unit digit 6? I mean it took me around 3 and half minutes to solve this question by the traditional way and that's way too long.

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

19 May 2020, 02:05

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sambitspm wrote:

Thanks GMATinsight. This is good trick, but what if we had two options ending in-unit digit 6? I mean it took me around 3 and half minutes to solve this question by the traditional way and that's way too long.

sambitspm

It means we are running out of luck

Another chance is to add them backwards ... That's lengthy but I am sure it won't take more than 3 minutes (I had problem recalling only 29^2 here)

31^2 = 961

29^2 = 841

27^2 = 729

25^2 = 625

23^2 = 529

21^2 = 441

19^2 = 361

17^2 = 289

15^2 = 225

13^2 = 169

11^2 = 121

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

24 May 2020, 09:15

Bunuel wrote:

What is the value of \(1^2 + 3^2 + 5^2 + ... + 31^2\)?

A. 9,455
B. 5,456
C. 4,892
D. 4,792
E. 3,468

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Asked: What is the value of \(1^2 + 3^2 + 5^2 + ... + 31^2\)?

\(t_n = (2n-1)^2 = 4n(n-1) + 1 = \frac{4}{3}(n-1)n(n+1) - \frac{4}{3}(n-2)(n-1)n + 1\)
\(S_n = \frac{4}{3}(n-1)n(n+1) + n = \frac{4}{3}* 15*16*17 + 16 = 5456\)

IMO B

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

25 May 2020, 06:50

GMATinsight wrote:

Bunuel wrote:

What is the value of \(1^2 + 3^2 + 5^2 + ... + 31^2\)?

A. 9,455
B. 5,456
C. 4,892
D. 4,792
E. 3,468

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Are You Up For the Challenge: 700 Level Questions

CONCEPT: Sum of n squares, \(∑n^2 = (\frac{1}{6})n(n+1)*(2n+1)\)

\(1^2 + 2^2 + 3^2 + ... + 31^2 = (\frac{1}{6})31(31+1)*(2*31+1) = (\frac{1}{6})31(32)*(63) = 31*16*21 = 10416\)

Now, subtract \(2^2 + 4^2 + ....+30^2 = 2^2*(1^2+2^2+3^2+....+15^2) = 4*(1/6)*15*16*31 = 4960\)

ie. \(1^2 + 3^2 + 5^2 + ... + 31^2 = 10416 - 4960 = 5456\)

ANswer: Option B

GMATinsight
what are you subtracting and why ? Isnt one formula enough to calculate the value

why this math is such a greedy subject

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

25 May 2020, 06:56

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dave13 wrote:

GMATinsight wrote:

Bunuel wrote:

What is the value of \(1^2 + 3^2 + 5^2 + ... + 31^2\)?

A. 9,455
B. 5,456
C. 4,892
D. 4,792
E. 3,468

Project PS Butler

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Are You Up For the Challenge: 700 Level Questions

CONCEPT: Sum of n squares, \(∑n^2 = (\frac{1}{6})n(n+1)*(2n+1)\)

\(1^2 + 2^2 + 3^2 + ... + 31^2 = (\frac{1}{6})31(31+1)*(2*31+1) = (\frac{1}{6})31(32)*(63) = 31*16*21 = 10416\)

Now, subtract \(2^2 + 4^2 + ....+30^2 = 2^2*(1^2+2^2+3^2+....+15^2) = 4*(1/6)*15*16*31 = 4960\)

ie. \(1^2 + 3^2 + 5^2 + ... + 31^2 = 10416 - 4960 = 5456\)

ANswer: Option B

GMATinsight
what are you subtracting and why ? Isnt one formula enough to calculate the value

why this math is such a greedy subject

Hey dave13

I am sure you have missed looking at question stem carefully.

Teh question is asking the sum of squares of ONLY ODD integers from 1 to 31

While the property gives us sum of squares of all (EVEN and ODD) integers therefore we found sum of all and subtracted sum of Squares of Even integers only

and yes, Math's is too greedy. Thank god... we teachers have something to contribute...

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

25 May 2020, 08:11

GMATinsight wrote:

dave13 wrote:

GMATinsight wrote:

Bunuel wrote:

What is the value of \(1^2 + 3^2 + 5^2 + ... + 31^2\)?

A. 9,455
B. 5,456
C. 4,892
D. 4,792
E. 3,468

Project PS Butler

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Are You Up For the Challenge: 700 Level Questions

CONCEPT: Sum of n squares, \(∑n^2 = (\frac{1}{6})n(n+1)*(2n+1)\)

\(1^2 + 2^2 + 3^2 + ... + 31^2 = (\frac{1}{6})31(31+1)*(2*31+1) = (\frac{1}{6})31(32)*(63) = 31*16*21 = 10416\)

Now, subtract \(2^2 + 4^2 + ....+30^2 = 2^2*(1^2+2^2+3^2+....+15^2) = 4*(1/6)*15*16*31 = 4960\)

ie. \(1^2 + 3^2 + 5^2 + ... + 31^2 = 10416 - 4960 = 5456\)

ANswer: Option B

GMATinsight
what are you subtracting and why ? Isnt one formula enough to calculate the value

why this math is such a greedy subject

Hey dave13

I am sure you have missed looking at question stem carefully.

Teh question is asking the sum of squares of ONLY ODD integers from 1 to 31

While the property gives us sum of squares of all (EVEN and ODD) integers therefore we found sum of all and subtracted sum of Squares of Even integers only

and yes, Math's is too greedy. Thank god... we teachers have something to contribute...

GMATinsight thanks

got it. just one question why are multiplying 1/6 by 4 ?

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

25 May 2020, 08:42

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Expert Reply

dave13 wrote:

GMATinsight wrote:

dave13 wrote:

GMATinsight wrote:

Bunuel wrote:

What is the value of \(1^2 + 3^2 + 5^2 + ... + 31^2\)?

A. 9,455
B. 5,456
C. 4,892
D. 4,792
E. 3,468

Project PS Butler

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Are You Up For the Challenge: 700 Level Questions

CONCEPT: Sum of n squares, \(∑n^2 = (\frac{1}{6})n(n+1)*(2n+1)\)

\(1^2 + 2^2 + 3^2 + ... + 31^2 = (\frac{1}{6})31(31+1)*(2*31+1) = (\frac{1}{6})31(32)*(63) = 31*16*21 = 10416\)

Now, subtract \(2^2 + 4^2 + ....+30^2 = 2^2*(1^2+2^2+3^2+....+15^2) = 4*(1/6)*15*16*31 = 4960\)

ie. \(1^2 + 3^2 + 5^2 + ... + 31^2 = 10416 - 4960 = 5456\)

ANswer: Option B

GMATinsight
what are you subtracting and why ? Isnt one formula enough to calculate the value

why this math is such a greedy subject

Hey dave13

I am sure you have missed looking at question stem carefully.

Teh question is asking the sum of squares of ONLY ODD integers from 1 to 31

While the property gives us sum of squares of all (EVEN and ODD) integers therefore we found sum of all and subtracted sum of Squares of Even integers only

and yes, Math's is too greedy. Thank god... we teachers have something to contribute...

GMATinsight thanks

got it. just one question why are multiplying 1/6 by 4 ?

\(2^2\) is outside the bracket and \((1/6)*15*16*31\) is the sum of squares from 1 to 15

B

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

25 May 2020, 10:10

sambitspm wrote:

Thanks GMATinsight. This is good trick, but what if we had two options ending in-unit digit 6? I mean it took me around 3 and half minutes to solve this question by the traditional way and that's way too long.

Most of the answers have different units digits, which means it has a high chance of working.

One note though is that you can move from one square to the next by adding:

\(1^2+1+2=2^2\)
\(2^2+2+3=3^2\)
\(3^2+3+4=4^2\) ... etc.

We can use this to work through the values of these squares, depending on how many of them we have memorized. Let's say we know all the squares to 20. Then:

\(21^2=20^2+20+21=441\)
\(23^2=21^2+21+22+22+23=441+88=529\) ... etc.

I'm not recommending doing the problem this way. This idea can help in certain niche situations, but is awkward here with a spacing of 2 between the squares, and so many of them.

B

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What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

01 Jun 2020, 11:05

To start with, I was not aware of the formula that exists for ∑n2.
So I just did the same problem in a slightly diff way using sequences
1^2+3^2+5^2+7^2+9^2+11^2+13^2+.........+21^2+23^2+.......+31^2
= 1^2+3^2+5^2+7^2+9^2+ (10+1)^2+(10+3)^2+..........+(20+1)^2+(20+3)^2+........+31^2

1^2+3^2+5^2+7^2+9^2= 165

(10+1)^2+(10+3)^2+(10+5)^2+(10+7)^2+(10+9)^2 = 5*10^2 + 20(1+3+5+7+9) +(1^2+3^2+5^2+7^2+9^2 )
Similarly
(20+1)^2+(20+3)^2+(20+5)^2+(20+7)^2+(20+9)^2 = 5*20^2 + 40(1+3+5+7+9) +(1^2+3^2+5^2+7^2+9^2)

= 3*165 + 5*(100+400) + 25(20+40) = 495+2500+1500 = 4495

Just add the LEFT OUT NUMBER 31^2; to this and the answer is there, 4495 + 961 = 5456

PS : pardon my ignorance of mathematical signs, inspite of detailed instructions on how to use the mathematical sign, i just couldn't get it working here. My first post, and its taken me 45 minutes to just type this......

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

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Re: What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2? [#permalink]

25 Sep 2021, 23:22

GMAT Problem Solving (PS) Questions

What is the value of 1^2 + 3^2 + 5^2 + ... + 31^2?