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

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: \(37^2\neq{36^2+1}\), \(38^2\neq{36^2+2}\), ... .

Hope it's clear.

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

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.

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.

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

(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

Bumping for review and further discussion.

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

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

For example, 17^2+23^2=818, but 7^2+3^2=58.

Hope it's clear.

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

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

Option C.
Using the formula for calculating sum of squares of n natural nos.:[n(n+1)(2n+1)]/6
First calculate for n=44 then for n=35
And subtract second from first to get the answer.

Posted from my mobile device

­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?­