What is the sum of all Perfect Squares lying b/w 3000 and 8000?

Question

any general method for this type of question? -thanks.

Answer 1

The general method is to:
(a) create an equation for sum of squares up to some number, call it s(n).
(b) subtract that equation for squares under 3,000, from the value for squares under 7,000, to eliminate the ones less than 3,000.

The sum of squares up to n^2 is:

1 + 4 + ... + n^2 = n(n+1)(2n+1)/6

So, since sqrt(3000) = ~54.77, the sum up to n=54 would be all perfect squares less than 3000.

And since sqrt(7000) = ~83.67, the sum up to n=83 would be all perfect squares less than 7000.

s(83) = 83 * (83 + 1) * (2*83 + 1) / 6 = 194,054

s(54) = 54 * (54 + 1) * (2*54 + 1) / 6 = 53,955

s(83) - s(54) = 140,099. That's the sum of all perfect squares beteen 3,000 and 7,000.

Answer 2

General method for this type of question:
1). Find the largest square less than 3000.
It is 54² = 2916 since 55² = 3025.
2). Find the largest square less than 8000.
It is 89² = 7921, since 90² = 8100.
3). Get a formula for the sum of the first n squares.
You can find it on Wikipedia, I'm sure.
It is n(n+1)(2n+1)/6.
4). Subtract the sum of all k² from k = 1 to 54
from the sum of all k² from k = 1 to 89.
5). Let S be your sum. Then the above "recipe"
gives
S = ( (89*90*179) - (54*55*109) )/6
= 185010.
Hope that helps!
BTW: McFate's work is correct, but he misread
8000 as 7000.

Answer 3

the square route of 3000 is 54.77 so we know that the first perfect square that lies in the given range has to be 55 because any integer less than 54.77 will be less than 3000 when squared. Similarly, the square route of 8000 is 89.44, so any integer greater than 89.44 will be outside the given range when squared. Therfore, the perfect squares that lie b/w 3000 and 8000 range from 55 to 89. The sum of these is 2520 I believe...

edit: 2520 is the sum of the square routes of the perfect squares so its not the answer you're looking for, but you can get the right answer by squaring and adding each number 55-89