Simple. You square the numbers and then calculate their sum - precisely as the name suggests.
So the answer is 1^2 + 4^2 + 2^2 + 5^2 = 1 + 16 + 4 + 25 = 46
To get a list of the squares of the first 1000 numbers we can do:> [n^2 | n sum [n^2 | n
If the regression sum of squares is the explained sum of squares. That is, the sum of squares generated by the regression line. Then you would want the regression sum of squares to be as big as possible since, then the regression line would explain the dispersion of the data well. Alternatively, use the R^2 ratio, which is the ratio of the explained sum of squares to the total sum of squares. (which ranges from 0 to 1) and hence a large number (0.9) would be preferred to (0.2).
The formula for the sum of the squares of odd integers from 1 to n is n(n + 1)(n + 2) ÷ 6. EXAMPLE : Sum of odd integer squares from 1 to 15 = 15 x 16 x 17 ÷ 6 = 680
The sum of the squares of the first 20 natural numbers 1 to 20 is 2,870.
The only squares of perfect squares in that range are 1, 16, and 81.
It is Fermat's theorem on the sum of two squares. An odd prime p can be expressed as a sum of two different squares if and only if p = 1 mod(4)
#include <iostream> using namespace std; int main() { int i,sum; // variables sum = 0; // initialize sum /* recursive addition of squares */ for (i = 1; i <= 30; i++) sum = sum + (i * i); cout << sum <<" is the sum of the first 30 squares." << endl; return 0; }
int sum (int n){if (n
The sum of the squares of the digits of 13 is 12 + 32 = 10. The sum of the squares of the digits of this result is 12 + 02 = 1. Because this process results in a 1, this number is a happy number.
The sum of the squares of the first 100 natural numbers [1..100] is 338350, while the sum of the first 100 natural numbers squared is 25502500.
12 + 42 + 82 = 81
If the sum of the squares of the vector's components is ' 1 ',then the vector's magnitude is ' 1 '.