What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ?

Question

What is the value of  38^2 + 39^2 + 40^2 + 41^2 + 42^2  ?

A. 7950
B. 7990
C. 8010
D. 8050
E. 8070

BrentGMATPrepNow · Accepted Answer

I love this question (and all of the replicas

)!! First notice that 38² + 39² + 40² + 41² + 42² = (40 - 2)² + (40 - 1)² + 40² + (40 + 1)² + (40 + 2)² Now notice that (40 - 2)² + (40 - 1)² + 40² + (40 + 1)² + (40 + 2)² has the form (x - 2)² + (x - 1)² + x² + (x + 1)² + (x + 2)² So, let's expand and simplify (x - 2)² + (x - 1)² + x² + (x + 1)² + (x + 2)² and see what we get. (x - 2)² + (x - 1)² + x² + (x + 1)² + (x + 2)² = + + + + = 5x² + 10 So, if (x - 2)² + (x - 1)² + x² + (x + 1)² + (x + 2)² simplifies to become 5x² + 10... ...then (40 - 2)² + (40 - 1)² + 40² + (40 + 1)² + (40 + 2)² must simplify to become 5(40²) + 10 5(40²) + 10 = 5(1600) + 10 = 8000 + 10 = 8010 Answer: C Cheers, Brent

Iamnowjust · Answer

My method:

(30+8) + (30+9) + (30+10) + (30+11) + 30+12) . I just did 8^2+9^2+10^2+11^2+12^2 = ...10, so the number should end with 10. The answer is C.

Posted from my mobile device

Bunuel · Answer

This is a copy of this question: https://gmatclub.com/forum/36-126078.html and this one: https://gmatclub.com/forum/what-is-the- ... 98546.html

chetan2u · Answer

It is always good to know the formula for SUM of square of first n natural numbers..  \frac{n(n+1)(2n+1)}{6} ..

so  38^2 + 39^2 + 40^2 + 41^2 + 42^2  = ( 1^2+2^2+............. 41^2 + 42^2 )-( 1^2 + 2^2 + 3^2 +...........+ 36^2 + 37^2 )..

\frac{42(42+1)(2*42+1)}{6}-\frac{37(37+1)(2*37+1)}{6}=7*43*85-37*19*25=8010

GMATBusters · Answer

I don't think this is right approach, this approach could have worked if we want to eliminate some options which do 
 not end with 0. here all the options end with 0. so this approach might not work.

Experts view needed  Bunuel , @chetan4u

GMATBusters · Answer

I really like your approach using this formula.
In fact I want to add 2 more fundamental formula for summation:

1)  1+2+3+...+n = n(n+1)/2  - Sum of first n natural numbers.
2)  1^3+2^3+...+n^3 = (n(n+1)/2)^2  - Sum of cube of first n natural numbers.

chetan2u · Answer

You are lucky to have got the correct answer..

There is no logic behind this..
example 38^2+39^2=2965, ...so __65
8^2+9^2=145

so you can just talk of UNITS digit and NOT tens digit

Funsho84 · Answer

(40-2)^2 + (40-1)^2 + 40^2 + (40+1)^2 + (40+2)^2

You can foil each expression for the answer. A quick way to the answer is to focus on the last multiplication when you foil for each expression.

You will get 4+1+4+1 = 10

Check out the link that Bunuel provided. His/her approach to this problem is really good.
https://gmatclub.com/forum/36-126078.html

dbushkalov · Answer

What if we are trying to find the sum of the squares of 6 or 7 consecutive integers? Is there a fixed formula for them too?

CEdward · Answer

I think you would just expand on Brent's formula?

Laveshan · Answer

Hi Guys, the long formulas scare me. I made a short cut formula based on Bunuels formulas:

A * B^2 + 2(C^2 + (C-1)^2 + (C-2)..until 0) = answer

A is how many numbers you need to add - 5
B is the middle number - 40 
C is the distance between the middle and lower (or higher number) - 2

We get: 5 * 40^2 + 2(2^2 + 2^1) = 8010.

I used this shortcut on other similar problems to test and it works well.

Bunuel, Karishma, Chetan and Brent, thank you for all the posts and work. It is really helping me!

adityasuresh · Answer

What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ?

We can use the common elements in (a+b)^2 and (a-b)^2
A^2 is going to 40 in all 5 elements: Therefore 40^2 * 5 = 8000
B^2 is going to be positive for all elements: Therefore 4+1+1+4
2AB Cancels out nicely with positive in 2ab in the first 2 elements and negative 2ab in the last 2 elements
Therefore sum = 8000 + 10
 C

BrushMyQuant · Answer

We need to find the sum of  38^2+39^2+40^2+41^2+42^2

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

38^2+39^2+40^2+41^2+42^2  =  1^2 + 2^2 + .... + 36^2+37^2+38^2+39^2+40^2+41^2+42^2  - ( 1^2 + 2^2 + .... + 36^2+37^2 ) 
 = Sum of Squares from 1 to 42 - Sum of Squares from 1 to 37
 =  \frac{42*(42+1)*(2*42+1)}{6}  -  \frac{37*(37+1)*(2*37+1)}{6} 
=  \frac{42*43*85}{6}  -  \frac{37*38*75}{6} 
= 25,585 - 17,575
= 8010

So,  Answer will C 
Hope it helps!

Watch the following video to MASTER Sequence problems

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

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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ?

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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ?

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