(n/2) times (n + 1)
eg for first 6 integers n = 6 and you have 6/2 x 6 + 1 = 3 x 7 = 21 which is correct.
13
No, there is not. Given any positive integer n, n+1 is also a positive integer and it is larger.
26
Sn = n*(n+1)
10 + 25n, where n is an integer.
The sum of any integer ( n ) and zero is ( n ).
Let's use N to represent any number.N x N = NN x -N = -N-N x -N = NSo the rules are:A positive integer times a positive integer will be a positive integerA positive integer times a negative integer will be a negative integerA negative integer times a negative integer will be a positive integer.
13
If the first integer is "n", then the second integer would be "n + 1", and the third, "n + 2". The sum of all this is (n) + (n + 1) + (n + 2) = 3n + 3. In other words, three times the first number in the sequence, plus 3.
No, there is not. Given any positive integer n, n+1 is also a positive integer and it is larger.
26
n2+n
The sum of the first forty positive integers can be calculated using the formula for the sum of an arithmetic series, which is (n/2)(first term + last term) where n is the number of terms. In this case, the sum is (40/2)(1 + 40) = 820.
Sn = n^2
Choose a nonzero integer for n to show -n can be evaluated as a positive number?
Sn = n*(n+1)
The formula for the first n positive numbers is n x (n+1)/2, so for n=11 the sum is 11 x 12 /2 = 66