mathematical proof question, please help me?

Question

prove that 1^3+2^3 +3^3 ..n^3 +(n+1)^3 = (1+2+3...+n+(n+1))^2

i really don't get it, thanks

Answer 1

This seems like something you want to prove by induction:

Base step (check that it holds for n=1):
n=1 => n+1=2
1^3+2^3=9=(1+2)^2

Now Assume that it's true for (n+1):
i.e. assume that:
1^3+2^3 +3^3 ..n^3 +(n+1)^3 = (1+2+3...+n+(n+1))^2

Using your assumption, try to show that it holds for n+2
i.e. want to show that:
1^3+2^3 +3^3 ..n^3 +(n+1)^3+(n+2)^3 = (1+2+3...+n+(n+1)+(n+2))^2
---------------
so we have:
1^3+2^3 +3^3 ..n^3 +(n+1)^3 + (n+2)^3
= (1+2+3...+n+(n+1))^2 + (n+2)^3
=(1+2+3...+n+(n+1))^2 + (n+2)^2*(n+2)
=(1+2+3...+n+(n+1))^2 + n(n+2)^2+2(n+2)^2
=(1+2+3...+n+(n+1))^2 + n(n^2+4n+4)+(n+2)^2+(n+2)^2
=(1+2+3...+n+(n+1))^2 + (n+1)(n+2)^2+(n+2)^2
=(1+2+3...+n+(n+1))^2 + (n+1)(n+2)(n+2)+(n+2)^2
=(1+2+3...+n+(n+1))^2 + n(n+1)(n+2)+2(n+1)(n+2)+ (n+2)^2
but now you're stuck, since you need to show the above equals the below:
=[1+2+3...+n+(n+1)]^2+2[1+2+3.... +n+(n+1)](n+2) + [n+2]^2
because you need to get:
(1+2+3...+n+(n+1)+(n+2))^2
from: (a+b)^2 = a^2+2ab+b^2)
where a=1+2+3...+n+(n+1) and b= (n+2)

therefore, we know we have to show that
(n+2)(n+1))=2(1+2+3..+n+n+1) by induction also
base case: (n=1)
(3)(2)=6=2(1+2)=2*3

induction hypothesis:
assume:
(n+2)(n+1))=2(1+2+3..+n+n+1)

induction:
wts: (n+3)(n+2))= 2(1+2+3..+n+n+1+n+2)

so
(n+3)(n+2)) = ([n+2]+1)([n+1]+1)= (n+1)(n+1)+(n+1)+(n+2)+1
=2(1+2+3..+n+n+1) + (n+1)+(n+2)+1
=2(1+2+3..+n+n+1) + 2n+4
=2(1+2+3..+n+n+1) + 2(n+2)
=2(1+2+3..+n+n+1+n+2)
so it's true, so we can go back and show that step above and our induction holds, so we proved it!

The reason we proved it is because we proved it for the first case, and if we assume it's true for any other case we proved it's true for the case after it. Which means that we proved it for all cases, since we can continually go to the next case forever!

Sorry it sucks that the algebra looks so messy, it'll be neater hopefully when you write it out on your paper. But the trick is that you have to do induction twice, otherwise you can't solve it. Although be careful, since sometimes you can't just do induction in the middle, sometimes you need to assume two things at once and prove them both at the same time. Good luck!

Answer 2

Induction

base case
n = 1
1^3 + 2^3 = 1 + 8 = 9
(1 + 2)^2 = 9

Assume true for n=k
Let LHS = 1^3+2^3 +3^3 ..k^3 +(k+1)^3
and RHS = (1+2+3...+k+(k+1))^2
Note: RHS = [(k+1)(k+2)/2]^2
= [(k+2)^2 * (k^2 + 2k + 1)] / 4
So [RHS / (k^2 + 2k + 1)] = [(k+2)^2]/4

Show true for n=k+1 such that:
1^3+2^3 +3^3 ..(k+1)^3 +((k+1)+1)^3 = (1+2+3...+(k+1)+((k+1)+1))^2

New LHS:
1^3+2^3 +3^3 + ... + k^3 + (k+1)^3 +((k+1)+1)^3
= 1^3 + 2^3 + ... + k^3 + (k + 2)^3 + (k + 1)^3
= 1^3 + 2^3 + ... + k^3 + k^3 + 6k^2 + 12k + 4 + (k + 1)^3
= LHS + k^3 + 6k^2 + 12k + 4

New RHS:
(1+2+3...+k+(k+1)+(k+2))^2
= [(k+2)(k+3) / 2]^2
= [(k + 2)^2 * (k^2 + 6k + 9)] / 4
= [RHS / (k^2 + 2k + 1)] * (k^2 + 6k + 9)

I'll continue this later, but you can see where I'm going.

Answer 3

I do not know how to do this directly, however it is rather straight forward to show by induction that:

1+2+3...+n = n(n+1)/2

and

1^3+2^3 +3^3 ..n^3 ={ [n(n+1)]^2}/4

So that gives you that the square of the first series is equal to the second.

I would be very interested in seeing if anyone knows of a direct proof.

Answer 4

The sum of cubes happens to be (1/4)n²(n+1)², which is the square of the sum of numbers (1/2)n(n+1). Derivation of the latter is trivial, but derivation of the former takes a little more work. However, here's a link offering a visual explanation for your problem.