8c cubed
Of the natural numbers, the smallest (excluding 0 and 1 which are the same to any power) cubes are: 8 (2 cubed), 27 (3 cubed), 64 (4 cubed), and 125 (5 cubed).
x times x times x equals x cubed
64 should be between 27 and 125. (1 cubed equals 1. 2 cubed equals 8. 3 cubed equals 27. 4 cubed equals 64. Etc.)
1 Sum of first n natural numbers = n(n+1)2[Formula.]2 Arthmetic mean of first n natural numbers = Sum of the numbers n[Formula.]3 = n(n+1)2n = n+124 So, the Arthmetic mean of first n natural numbers = n+12
N cubed
n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
n3 - 93 can be factorised into (n - 9)(n2 + 9n + 81)
n^3 (n cubed)
2 N cubed (3)
The sum of the first n cubed numbers is: [n*(n+1)/2]2 which is the same as the square of the sum of the first n numbers.
"n cubed" is not an algebra problem, since it asks no question. It's an "expression", whose numerical value depends on the value of ' n '.
. N times 6 cubed
It is n cubed.
6n5
Firstly to get 5n you times by 5!Next n3 (n cubed) is n x n x nTherefore you are timesing n by 5 and then by n and nSo 5n3 = times n by 5 x n x n or simplifyed n x 5n2
The sum of the first n cubed numbers is the square of the nth triangular number.