why does the sum of consecutive cube numbers starting from 1 add up to the square of the uncubed number
e.g. 1 cube + 2 cube = (1+2) square
1 cube + 2 cube + 3 cube + 4 cube + 5 cube = (1+2+3+4+5) square
use algebra to explain how it is such?
maths is beautiful
This is true. Here is the Algebra involved:
Solve for x over the real numbers:
x^3+(x+1)^3 = (2 x+1)^2
Expand out terms of the left hand side:
2 x^3+3 x^2+3 x+1 = (2 x+1)^2
Expand out terms of the right hand side:
2 x^3+3 x^2+3 x+1 = 4 x^2+4 x+1
Subtract 4 x^2+4 x+1 from both sides:
2 x^3-x^2-x = 0
The left hand side factors into a product with three terms:
x (x-1) (2 x+1) = 0
Split into three equations:
x-1 = 0 or x = 0 or 2 x+1 = 0
Add 1 to both sides:
x = 1 or x = 0 or 2 x+1 = 0
Subtract 1 from both sides:
x = 1 or x = 0 or 2 x = -1
Divide both sides by 2:
Answer: |
| x = 1 or x = 0 or x = -1/2
+130540 We can prove this by mathematical induction......
First.....the sum of the first n positive integers = [n][n+1]^2/ 2
Therefore, the [sum of the first n positive integers]^2 = [n]^2 [n =1]^2 / 4
So....we want to prove that
1^3 + 2 ^3 + 3^3 + .....+ n^3 = [n]^2 [n =1]^2 / 4
Show that it is true for n = 1
1^3 = [1]^2 [2)^2 / 4 = [1][4]/4 = 1
Assume it's true for k.....that is......
1^3 + 2^3 + 3^3 + ....... + k^3 = [k]^2 [k + 1]^2/ 4
Show that it is true for k + 1 .... that is.........
1^3 + 2^3 + 3^3 + .......+ k ^3 + [k + 1]^3 = [k + 1]^2 [k + 2]^2 / 4
So we have.......
1^3 + 2^3 + 3^3 + .......+ k ^3 + [k + 1]^3 = [k]^2 [k + 1]^2 / 4 + [k + 1]^3
1^3 + 2^3 + 3^3 + .......+ k ^3 + [k + 1]^3 = [k + 1]^2 ( k^2 + 4[k + 1] ) / 4
1^3 + 2^3 + 3^3 + .......+ k ^3 + [k + 1]^3 = [k + 1]^2 ( k^2 + 4k + 4 ) / 4
1^3 + 2^3 + 3^3 + .......+ k ^3 + [k + 1]^3 = [k + 1]^2 [k + 2]^2 / 4
Which is what we wanted to prove
