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.

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)

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.