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The sum of the squares of the first 15 positive integers

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by abhi332 » Thu Feb 25, 2010 1:12 pm
The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the
sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ?
(A) 2480
(B) 3490
(C) 6785
(D) 8215
(E) 9255

[spoiler]OA:D[/spoiler]
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by harsh.champ » Thu Feb 25, 2010 1:46 pm
abhi332 wrote:The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the
sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ?
(A) 2480
(B) 3490
(C) 6785
(D) 8215
(E) 9255

[spoiler]OA:D[/spoiler]
Sum of squares of n natural no.s = n(n+1)(2n+1)/6
Now,sum upto 15 = 15(15+1)(2(15)+1)/6 -(1)
Also,sum upto 30 = 30(30+1)(2(30)+1)/6 -(2)
Thus,sum from 16 to 30 can be found from subtracting (1) from (2),
we get the answer as 8215

I solved by this method in about 2 mins.
Last edited by harsh.champ on Thu Feb 25, 2010 1:57 pm, edited 1 time in total.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by harsh.champ » Thu Feb 25, 2010 1:51 pm
Alternative shorter approach
abhi332 wrote:The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the
sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ?
(A) 2480
(B) 3490
(C) 6785
(D) 8215
(E) 9255

[spoiler]OA:D[/spoiler]
Now,since we are given the sum of squares of 1st 15 no.s,so why not use that to ease our our calculation.
[Just check the 2 formulas made in the above post]
From (1) we have, (15)(16)(31)/6 = S 15 =1240
From (2) we have , (30)(31)(61)/6 = S 30 = x
Dividing (2) by (1),we get
(2*61)/16 = x/1240
x=9455
[spoiler]hence, sum =9455-1240=8215= ( D )[/spoiler]

Solving by this approach saves atleast 1min. of your time (I solved by this way in around 40 seconds)
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by GMATGuruNY » Wed Jul 09, 2014 3:08 pm
abhi332 wrote:The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the
sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ?
(A) 2480
(B) 3490
(C) 6785
(D) 8215
(E) 9255
One of my students today asked how to solve this problem, so I thought I'd share my approach.

16² = (15 + 1)² = 15² + 2*15*1 + 1² = 15² + 30 + 1².
17² = (15 + 2)² = 15² + 2*15*2 + 2² = 15² + 60 + 2².
18² = (15 + 3)² = 15² + 2*15*3 + 3² = 15² + 90 + 3².

Notice the pattern.
The red terms are all 15².
The blue terms are consecutive multiples of 30.
The green terms are consecutive perfect squares, starting with 1².

When we add together the perfect squares between 16² and 30², inclusive -- for a total of 15 values -- the sum will be composed of 15 red terms, 15 blue terms, and 15 green terms.

Red terms:
15 red terms = 15 * 15² = 15 * 225 = 10*225 + 5*225 = 2250 + 1125 = 3375.

Blue terms:
The first 15 multiples of 30 are {30, 60, 90....450}.
For any evenly spaced set:
Average = (biggest + smallest)/2.
Sum = (number of terms)(average).
For the 15 blue terms:
Average = (450+30)/2 = 240.
Sum = 15*240 = 10*240 + 5*240 = 2400 + 1200 = 3600.

Green terms:
The problem indicates that the sum of the first 15 positive perfect squares = 1240.

Resulting sum:
red terms + blue terms + green terms = 3375 + 3600 + 1240 = 8215.

The correct answer is D.
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