(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
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)
First, an equation is needed. We will designate the first integer as N, and the second as N + 2. The sum of the two numbers is 176, so the equation should look like this:N + N + 2 = 176Now it needs to be added:2N + 2 = 176Now both sides need to be subtracted by 2:2N = 174Now both sides need to be divided by 2:N = 87Therefore, the first integer is 87. Since the second integer is designated by N + 2, and N is equal to 87, the second integer is 87 plus 2, which is equal to 89.So the two consecutive odd integers whose sum is 176, are 87 and 89.
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.
26
No, there is not. Given any positive integer n, n+1 is also a positive integer and it is larger.
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 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