36^2 + 37^2 + 38^2 + 39^2 + 40^2 + 41^2 + 42^2 + 43^2 + 44^2 =

Question

36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=

(A) 14400
(B) 14440 
(C) 14460
(D) 14500
(E) 14520

Guys this is how I am trying to solve this. But after doing it for 15 minutes I gave up and think to post. No idea where I am getting this wrong and i don't have an OA either.

Here we go:

The question can be simplified as:

36^2 +{36+1}^2 +{36+2}^2 +{36+3}^2{36+4}^2{36+5}^2{36+6}^2{36+7}^2{36+8}^2

The above can be further simplified into

36^2  + {1296+1+72} + {1296+4+44} + ..................... + {1296+64+576} ------->  (a+b)^2  formula] -------------------------------------------(1)

Now the (1) can be simplified into
1296 {1+74+149+226+.................+641} ==> 1296*2805 => This doesn't give me any of the above choices. Any idea guys where I am getting it wrong?

Bunuel · Accepted Answer

Approach #1:

We have 9 terms, middle term is 40^2. Express all other terms as 40-x:
 36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2= 
 =(40-4)^2+(40-3)^2+(40-2)^2+(40-1)^2+40^2+(40+1)^2+(40+2)^2+(40+3)^2+(40+4)^2 . Now, when you expand these expressions applying  (x-y)^2=x^2-2x+y^2  and  (x+y)^2=x^2+2x+y^2  you'll see that  -2xy  and  2xy  cancel out and we'll get:

(40^2+4^2)+(40^2+3^2)+(40^2+2^2)+(40^2+1^2)+40^2+(40^2+1^2)+(40^2+2^2)+(40^2+3^2)+(40^2+4^2)= 
 =9*40^2+2*(4^2+3^2+2^2+1^2)=14,400+60=14,460

Approach #2:  
The sum of the squares of the first n positive integers is given by:  1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}  (note that it's highly unlikely that you'll need it on the real GMAT test). For example the sum of the first 3 positive integers:  1^2+2^2+3^3=\frac{3(3+1)(2*3+1)}{6}=14 .

Now, we can calculate the sum of the squares of the first 44 positive integers and subtract from it the sum of the squares of the first 35 positive integers to get the answer:  36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=\frac{44(44+1)(2*44+1)}{6}-\frac{35(35+1)(2*35+1)}{6}=14,460 .

Answer: C.

Hope it helps.

LalaB · Answer

hm, before looking to ur solution, I solved the q. in this way-
let a =36^2
then we have a+(a+1)+(a+2) (a+3) (a+4) (a+5) (a+6) (a+7) (a+8)

I feel that it is just an arithmetic progression with mean=median

so the sum of these numbers are 9*(a+4)=9*40^2=14400

later I saw ur solution - =9*40^2+2*(4^2+3^2+2^2+1^2)=14,400+60=14,460 
we seem to be in the same way, but using my method how to come to +2*(4^2+3^2+2^2+1^2)?

Bunuel · Answer

The problem is that 36^2, 37^2, 38^2, 39^2, 40^2, 41^2, 42^2, 43^2, and 44^2 DOES NOT form an AP. You assumed that  a+1=36^2+1=37^2 ,  a+2=36^2+2=38^2 , ... but that's not correct:  37^2
eq{36^2+1} ,  38^2
eq{36^2+2} , ... .

Hope it's clear.

LalaB · Answer

Bunuel, yep, u r right. my idea was wrong. thnx

IanStewart · Answer

It is not an arithmetic progression, so if you assume that it is, you will get the wrong answer. An arithmetic progression is 'equally spaced'. If you just look at the smallest perfect squares, you can see that they are not equally spaced: 1, 4, 9, 16, 25... In fact the spacing gets larger the further you get into this list.

You can use the spacing of perfect squares to answer this question. From the difference of squares, we have that

41^2 - 40^2 = (41 + 40)(41 - 40) = 81

So 41^2 = 40^2 + 81. Similarly, 42^2 = 41^2 + 83, and 39^2 = 40^2 - 79, and so on. Listing all of the values we need to sum:

\begin{align*}
36^2 &= 40^2 - 79 - 77 - 75 - 73\
37^2 &= 40^2 - 79 - 77 - 75 \
38^2 &= 40^2 - 79 - 77 \
39^2 &= 40^2 - 79 \
40^2 &= 40^2 \
41^2 &= 40^2 + 81 \
42^2 &= 40^2 + 81 + 83 \
43^2 &= 40^2 + 81 + 83 + 85 \
44^2 &= 40^2 + 81 + 83 + 85 + 87 
\end{align*}

Now adding these in columns, we get (9)(40^2) + 4(81-79) + 3(83 - 77) + 2(85 - 75) + (87 - 73) = 9*1600 + 4*2 + 3*6 + 2*10 + 14 = 14,400 + 8 + 18 + 20 + 14 = 14,460

I'd still probably use the first method outlined in Bunuel's post above, but you can use the spacing of squares to get the answer if you look at the problem in the right way.

someone79 · Answer

It would be easier if you think about this problem in this way:

(40-4)^2 + (40-3)^2 + (40-2)^2 + (40-1)^2 + (40-0)^2 + (40+1)^2+ (40+2)^2 + (40+3)^2 + (40+4)^2

all the 4*40*2 or 2* a*b terms will get cancelled out.

1600*9 + (4*4 + 3*3 + 2*2 + 1)*2 = answer = 14460

any other way would be too lengthy.

take middle term in such cases and try to formulate it in ways such as (a-1)^2 + (a+1)^2.

BN1989 · Answer

I also used this approach, but it's really calculation intensive.

Apex231 · Answer

(A)14400
(B)14440 
(C)14460
(D)14500
(E)14520

Sum of squares = (n)(n+1) (2n+1) /6
Sum of squares (44) - Sum of squares (35)

(44 * 45 * 89) /6 – (35 * 36 * 71)/6

(176220 – 89460) / 6 = 86760 / 6 = 14460

Bunuel · Answer

Bumping for review and further discussion.

atalpanditgmat · Answer

Hi Bunuel,
I have different way but do not know it is a mere coincidence or valid approach. Is this logic ok?

36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2

Taking square of unit digits:
6^2= 36
7^2= 49
8^2=64
9^2=81
0^2=0
1^2=1
2^2=4
3^2=9
4^2=16

Adding all the values 36+49+64+81+1+4+9+16=  260

Only option C has 60 on the last. Hence the answer is C.

Atal Pandit

Bunuel · Answer

No, that's not correct. You can only get the units digit with this approach.

For example, 17^2+23^2=8 18 , but 7^2+3^2=58.

Hope it's clear.

PareshGmat · Answer

What level question is this please? Its taking more than 2 minutes for me

Bunuel · Answer

I'd say above 700, so spending extra 1 minute on those is OK.

AKG1593 · Answer

Option C.
Using the formula for calculating sum of squares of n natural nos.: /6
First calculate for n=44 then for n=35
And subtract second from first to get the answer.

Posted from my mobile device

tksvaan · Answer

Here's my approach.

sum of 9 consecutive numbers => 1 + 2 + 3 + 4 + 5 + 6 +7 + 8 + 9 => divided by 9 evenly
sum of 9 consecutive square numbers => 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 = (81+9) + (64+36) + (49+1) + (16+4) + 25 => divided by 9 remain 6

Only option C when divided by 9 give a remainder of 6.

I tried this with 6 consecutive numbers and still get the correct answer. Can you please check if my approach is correct?

BrushMyQuant · Answer

We need to find the sum of  36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2

We know that sum of first n positive integers starting from 1 =  \frac{n*(n+1)*(2n+1)}{6}

36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2  =  1^2 + 2^2 + .... + 36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2  - ( 1^2 + 2^2 + .... + 34^2+35^2 ) 
 = Sum of Squares from 1 to 44 - Sum of Squares from 1 to 35
 =  \frac{44*(44+1)*(2*44+1)}{6}  -  \frac{35*(35+1)*(2*35+1)}{6} 
=  \frac{44*45*89}{6}  -  \frac{35*36*71}{6} 
= 29,370 - 14,910
= 14,460

So,  Answer will C 
Hope it helps!

Watch the following video to MASTER Sequence problems

https://www.youtube.com/watch?v=EfNT2JoE5CM

bumpbot · Answer

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36^2 + 37^2 + 38^2 + 39^2 + 40^2 + 41^2 + 42^2 + 43^2 + 44^2 =

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36^2 + 37^2 + 38^2 + 39^2 + 40^2 + 41^2 + 42^2 + 43^2 + 44^2 =

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