n = number n + 6 ======
6 + n = 21
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.
6(n + y)
(6+n) n= a number
n + (8 + 6)
It is n + 6.
n = number n + 6 ======
the sum of 3 times m and n
6+n
6 + n = 21
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.
6(n + y)
mn = 8 ( Product) m + n = 6 (Sum) Hence m = 6 - n Substitute (6 -n)n = 8 Multiply out the brackets 6n - n^(2) = 8 n^(2) - 6n + 8 = 0 It is now in quadratic form and will factor Hence ((n - 2)(n - 4) = 0 n = 2 & n = 4 So '2' & '4' are the two numbers. Verification 2 + 4 = 6 2 x 4 = 8
6 + N where N is the unspecified number.
(6+n) n= a number
The formula for the sum of the squares of odd integers from 1 to n is n(n + 1)(n + 2) ÷ 6. EXAMPLE : Sum of odd integer squares from 1 to 15 = 15 x 16 x 17 ÷ 6 = 680