algebra
posted by jadacess .
The sum of the swuares of the first n counting numbers:
1² + 2² + 3² + 4² + 5² + 6² + ... +n²
is given by the formula
S = n(n+1)(2n+1)

6
Find the sum of the squares of all the counting numbers between 15 and 35.
 answer is 12445;
how do you do this question ? ><

Use the formula to find the sum of the squares from 1 to 35,
then find the sum of the squares of the numbers from 1 to 14
Subtract the smaller from the larger to get the inbetweens.
I had 14910  1015 = 13895
did you mean ... 15 to 35 inclusive or exclusive?
Mine is the inclusive, for the exclusive
subtract up to 15 from up to 34, that will give you the answer they provided. 
The sum of the swuares of the first n counting numbers:
1² + 2² + 3² + 4² + 5² + 6² + ... +n²
is given by the formula
S = [n(n+1)(2n+1]/6
The sum of the squares of the first "n" counting numbers minus the squares of the first "m  1) counting numbers (or the sum of the squares from "m" to "n")is given by the formula
S = [2(n^3  m^3) + 3(n^2  m^2) + (n  m)]/6.
The sum of the squares from n = 1 to n = 7 is S(17) = 7(8)(15)/6 = 140.
The sum of the squares from "m" = 4 to n = 7 is S(47) = [558 + 195 + 3]/6 = 126.
The sum of the squares of all the counting numbers between 15 and 35.
Do you mean the sum of squares of the counting numbers "between" 15 and 35, or 16 through 35 or the counting numbers "from" 15 through 35, or 15 through 35. You decide and use the formula.