There is no formula that will sum n even numbers without further qualifications: for example, n even numbers in a sequence.
The sum of two even numbers is always EVENYes.Proof:We can consider an even number as 2*n, and an odd number as 2*n+1.The sum of two even numbers would be 2*n+2*m=2*(n+m), an even number.
If n is the first number, n + n + 1 + n + 2 + n + 3 will be the sum, which is 4n + 6. 4n is always even, and 6 is even. Therefore, the sum of four consecutive numbers is always even.
n*(n+1)
1. start 2. sum=0 3. input n 3. for i=1 to n do 4. input x 5. sum=sum+x end of for (3) 6. avg=sum/n 7. output sum, avg 8. stop
sum = 0 for(n = 0; n <= 10; n += 2) sum += n;
TO find the sum of n numbers?
Oh, what a lovely question! To compute the sum of the squares of N numbers, you can create a simple algorithm. Start by initializing a variable to hold the sum, then loop through each number, square it, and add it to the sum. Once you've done this for all N numbers, you'll have the sum of their squares. Just like painting a happy little tree, take your time and enjoy the process.
Algorithm: sum_evenInput: an integer, n, such that n>0Output: the sum of all even integers within the open range (1:n)sum := 0val := 2while (val < n) do{sum := sum + valval := val + 2}return sumNote that you explicitly specified between 1 and n, which literally means both 1 and n should be excluded from the sum. 1 would be excluded anyway since it is not an even number, however if we wish to include n, then use the following algorithm instead:Algorithm: sum_evenInput: an integer, n, such that n>0Output: the sum of all even integers within the half-open range (1:n]sum := 0val := 2while (val
Sum = 0 For N = 1 to 10 Sum = Sum + 2*N Next N Print Sum
There is no formula that will sum n even numbers without further qualifications: for example, n even numbers in a sequence.
1.start 2.n=1,s=0 3 compute s=s+n 4 compute n=n+1 5 check n<=10 go to step3 else go to step 7 6 display s 7 stop
The sum of two even numbers is always EVENYes.Proof:We can consider an even number as 2*n, and an odd number as 2*n+1.The sum of two even numbers would be 2*n+2*m=2*(n+m), an even number.
If n is the first number, n + n + 1 + n + 2 + n + 3 will be the sum, which is 4n + 6. 4n is always even, and 6 is even. Therefore, the sum of four consecutive numbers is always even.
The general equation to find the sum of the numbers 1 to n is: (n*(n+1))/2So, for n=10, you have:(10*(10+1))/2(10*11)/2110/255
Sum = n/2[2Xa1+(n-1)d] where n is last number, a1 is the first number & d is the common difference between the numbers, here d=2 for the even /odd numbers. Sum = n/2 [2Xa1+(n-1)2]
Sum of 1st n even numbers: count*average = n * (2 + 2*n)/2 = n * (n+1) Sum = 50 * (2+100)/2 = 50*51 = 2550