13
26
No, there is not. Given any positive integer n, n+1 is also a positive integer and it is larger.
Sn = n*(n+1)
The sum of the first n positive integers (i.e. 1+2+3+...+n) is given by n(n+1)/2 so the sum of the first 25 positive integers is 1+2+3+...+25 = 25x26/2 = 325. The sum of the first 25 EVEN numbers will be double this i.e. 2+4+6+...+50 = 650.
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
Sn = n*(n+1)
Choose a nonzero integer for n to show -n can be evaluated as a positive number?
The sum of the first n positive integers (i.e. 1+2+3+...+n) is given by n(n+1)/2 so the sum of the first 25 positive integers is 1+2+3+...+25 = 25x26/2 = 325. The sum of the first 25 EVEN numbers will be double this i.e. 2+4+6+...+50 = 650.