The total is 465
The sum of whole numbers 1 through 30 is 465.
(30 + 1) * 30/2 ie 465
The product of a sum and a difference with 31 times 29 is:(30+1) x (30-1)sum to differencemakea product= 30² - 1²= 900 - 1= 899
465
1+30=31 and 1x30=30
The sum of whole numbers 1 through 30 is 465.
#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; }
(30 + 1) * 30/2 ie 465
The product of a sum and a difference with 31 times 29 is:(30+1) x (30-1)sum to differencemakea product= 30² - 1²= 900 - 1= 899
465
1+30=31 and 1x30=30
Oh, dude, math time! So, if 30% of the sum is 750, we can figure out the whole sum by doing some basic math. We just divide 750 by 0.30 to find out the total sum, which comes out to be 2500. So, the sum of money is 2500 bucks. Math made easy, like a piece of cake... or pie.
To total numbers from 1 through 100 using a FOR-NEXT loop, you would initialize a variable to store the sum and then iterate through each number from 1 to 100. In each iteration, you would add the current number to the sum variable. Here's a simple pseudocode example: sum = 0 FOR i = 1 TO 100 sum = sum + i NEXT i After the loop, the variable sum will contain the total of numbers from 1 to 100.
The factors of 3 are 1 and 3. The sum of the digits of these factors is calculated as follows: for 1, the sum of its digits is 1, and for 3, the sum of its digits is 3. Therefore, the total sum of the digits of the factors of 3 is 1 + 3 = 4.
3/10 + 1/3... = 9/30 + 10/30... = 19/30
Suppose the numbers are x and y. The sum of their reciprocals = 1/x + 1/y = y/xy + x/xy = (y+x)/xy = (x+y)/xy = 10/30 = 1/3
29 and 30.