Prove that the sum of qubes of the integers (i^3) from 1 to n is equal to:(1/4)n^2(n+1)^2?

Question

Thank you all the 'inductionists' what would you do if you did not know the answer? Can someone actually show how to construct this formula?

Answer 1

Let sum of cubes to n =

1/4 n^2 (n+1)^2 = 1/4 n^4 + 1/2 n^3 + 1/4 n^2

Add the next cube (n+1)^3 = n^3 + 3 n^2 + 3 n + 1, so that we have:

1/4 n^4 + 3/2 n^3 + 13/4 n^2 + 3 n + 1 =

1/4 (n^2 + 2n + 1) (n^2 + 4n + 4) =

1/4 (n+1)^2 (n+2)^2

QED, by induction.

Answer 2

The easiest way to prove it is by induction. That means:

1) Show it is true for n = 1
2) Show that if it is true for n, it is also true for n+1.

If the formula for the sum S, of the cubes 1 thru n is:

S = [n(n + 1)/2]^2

Then show that the sum T, of the cubes 1 thru (n+1) is

T = [(n + 1)(n + 2)/2]^2
_____________

For n = 1 we have

S = [n(n + 1)/2]^2 = [1*(1 + 1)/2]^2 = [1*2/2]^2 = 1^2 = 1
This checks out.

T = S + (n + 1)^3 = [n(n + 1)/2]^2 + (n + 1)^3
T = (n^4 + 2n^3 + n^2)/4 = (n^3 + 3n^2 + 3n + 1)
T = (n^4 + 2n^3 + n^2)/4 = (4n^3 + 12n^2 + 12n + 4)/4
T = (n^4 + 6n^3 + 13n^2 + 12n + 4)/4
T = [(n + 1)(n + 2)/2]^2

qed
by induction

Answer 3

You have to use mathematical induction procedure. Email me & I'll send you a PDF of the solution

Answer 4

dunno