By learning how to program on C+.
The first ten positive numbers total 55.
one
product = 6 first three positive counting numbers = 1, 2, 3product of the first three positive counting numbers:1 x 2 x 3 = 6
2,3,5,6,7,8
1,2,3,4,5
printf("1 3 5 7 9\n");
The first ten positive numbers total 55.
The sum of the first 30 positive even numbers is 930.
The sum of the first 100 positive even numbers is 10,100.
The sum of the first 30 positive even numbers is 930.
The sum of the first six positive numbers (1 to 6) is 21.
one
The sum of the first seven positive INTEGERS is 28. The sum of the fisrt seven positive numbers is infinitesimally small.
product = 6 first three positive counting numbers = 1, 2, 3product of the first three positive counting numbers:1 x 2 x 3 = 6
2,3,5,6,7,8
1,2,3,4,5
If you know that the product of 2 negative numbers is positive, then the product of 4 negative numbers has to be positive. The product of the first two negative numbers is positive and the next two negative numbers is positive. Multiplying the product of the first two numbers (positive number) and the product of the last two numbers (also positive), is a positive number times a positive number which is positive. Let a, b, c and d be negative numbers: (a*b*c*d) = (a*b)*(c*d) (-ve*-ve*-ve*-ve)=(-ve*-ve)*(-ve*-ve)= (+ve)*(+ve) = (+ve)