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

Question

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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gmatlawyer · Accepted Answer

1^2 + 3^2 + 5^2 + 7^2 + ... + 31^2

Observe that all the options have different units digits; therefore, we will just concentrate on the units digits of the squares of the numbers.

Units digits for the numbers are as follows:
  1^2=1

3^2=9

5^2=5

7^2=9

9^2=1

11^2=1

13^2=9

We see a pattern here for the units digits, which is 1, 9, 5, 9, 1 and then it repeats for every 5 counts.

So, from 1 through 31(inclusive), there are 16 odd digits, that means that the above pattern will repeat for 3 times and then the first digit will repeat for the 16th term i.e. 31. As such, we will add the digits of the pattern i.e.  1+9+5+9+1  which is  25 , multiple it by 3 times since the pattern repeats thrice, which is equal to  75 , and add the first digit of the pattern i.e.  75+1 . This gives us  76 .

We can conclude that the units digit of the sum of all the numbers is  6 ; which corresponds to option B.

GMATinsight · Answer

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

nick1816 · Answer

N=  -

N=  - 2^2

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

lacktutor · Answer

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).

sambitspm · Answer

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

chetan2u · Answer

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.

GMATinsight · Answer

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

sambitspm · Answer

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.

GMATinsight · Answer

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

Kinshook · Answer

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

dave13 · Answer

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

why this math is such a greedy subject

GMATinsight · Answer

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...

dave13 · Answer

GMATinsight thanks

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

GMATinsight · Answer

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

WheatyPie · Answer

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.

sreeni58 · Answer

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......

Kinshook · Answer

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

t_n = (2n-1)ˆ2 = 4nˆ2 - 4n + 1 
Number of terms = n = (31 - 1) / 2 + 1 = 16
 S_n = 4n(n+1)(2n+1)/6 - 4n(n+1)/2 + n = 4*16*17*33/6 - 4*16*17/2 + 16 = 5456

IMO B

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

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

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