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555-605 Level|   Arithmetic|      
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KSBGC
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? Updated on: 09 Apr 2018, 07:08
Originally posted by KSBGC on 09 Apr 2018, 06:55.
Last edited by Bunuel on 09 Apr 2018, 07:08, edited 1 time in total.
Renamed the topic and edited the question.
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C
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77% (02:15) correct 23% (02:16) wrong based on 758 sessions

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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
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C
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BrentGMATPrepNow
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 09 Apr 2018, 07:48
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selim
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

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 = [x² - 4x + 4] + [x² - 2x + 1] + [x²] + [x² + 2x + 1] + [x² + 4x + 4] [lots of canceling to do here...]
= 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
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 09 Apr 2018, 09:02
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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.

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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 09 Apr 2018, 07:06
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selim
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

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
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 09 Apr 2018, 08:43
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selim
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

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

\(\frac{42(42+1)(2*42+1)}{6}-\frac{37(37+1)(2*37+1)}{6}=7*43*85-37*19*25=8010\)
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 09 Apr 2018, 09:20
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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

Iamnowjust
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.

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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 09 Apr 2018, 09:24
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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
selim
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

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

\(\frac{42(42+1)(2*42+1)}{6}-\frac{37(37+1)(2*37+1)}{6}=7*43*85-37*19*25=8010\)
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 09 Apr 2018, 09:45
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Iamnowjust
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


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
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 09 Apr 2018, 10:28
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(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
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 29 Oct 2019, 14:21
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selim
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

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 = [x² - 4x + 4] + [x² - 2x + 1] + [x²] + [x² + 2x + 1] + [x² + 4x + 4] [lots of canceling to do here...]
= 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


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?
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 22 Nov 2020, 19:59
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selim
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

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 = [x² - 4x + 4] + [x² - 2x + 1] + [x²] + [x² + 2x + 1] + [x² + 4x + 4] [lots of canceling to do here...]
= 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


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?

I think you would just expand on Brent's formula?
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 23 Nov 2020, 04:40
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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!
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 12 Feb 2023, 07:59
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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
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What is the value of 38^2 + 39^2 + 40^2 + 41^2 + 42^2 ? 21 Sep 2024, 22:44
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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!

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