That is an unsolvable problem, because there are an infinite number of even numbers. However, with limits, you could say...
int NMax = ...;
int Sum = 0;
for (N=2; N<=NMax N+=2) Sum += N;
Count how many of the numbers are odd. If an odd number of them are odd, then their sum is odd. If an even number of them are odd, then their sum is even.
The sum of the first 10 even numbers is 110.
Even. The sum of two even numbers is always even.
There is no formula that will sum n even numbers without further qualifications: for example, n even numbers in a sequence.
int i, sum = 0; for (i=0; i<20; i+=2) sum+=i;
Because the sum of two odd numbers is always even and the sum of that even number plus another odd number is always odd.
No. The sum of four consecutive integers is two odd numbers plus two even numbers which is an even number. 2001 is an odd number, therefore it cannot be the sum of four consecutive numbers.
no
of all the things i got is 2550
The sum of two evens is even so the sum of any number of evens is even. It is, therefore, impossible for the sum of three even numbers, whether or not consecutive, to be 57, which is an odd number.
4 plus 10 plus 16 plus 70 equals 100. To find the sum of this series, simply add all the numbers together.
The numbers are 32, 34, and 36.