Current Events > Which of these mathematical values is greater?

(+) Yeah! (+)

First a warm up...

A: (1+2+3+...+99+100)^2
B: (1^2)+(2^2)+(3^2)+...+(99^2)+(100^2)

Basically, the square of the sum vs the sum of squares. Not too hard, right?

Now for the real problem...

A: (1+2+3+...+99+100)^2
B: (1^3)+(2^3)+(3^3)+...(99^3)+(100^3)

In this case, it's the square of the sum vs the sum of cubes.

For the real problem, A and B are equal.

The warm up can be done by inspection: A > B because A includes every term that is in B plus positive cross products.

Instinct tells me A both times. Looking forward to a math chad showing the work.

It turns out that this is a bit easier than I originally thought. Here's the answer showing the work (the topic is old enough that I'm not going to spoiler tag everything).

1) 100 here is a distraction -- let's look at the problem for any n, starting with n=2,3,4. We have

(1+2)^2 = 9 = 1 + 8 = 1^3 + 2^3

(1+2+3)^2 = 36 = 1 + 8 + 27 = 1^3 + 2^3 + 3^3.

(1+2+3+4)^2 =100 = 1 + 8 + 27 + 64 = 1^3 + 2^3 + 3^3 + 4^3.

2) This suggests that A and B are equal for all n. The proof can be done by direct computation.

3) Let S(n) be the sum 1+ 2 + ... + n = n + (n-1) + ... + 1

Adding these two expressions for S(n) term by term gives

2S(n) = (1+n) + (2+n-1) + (3+n-2) + ... + (n-1+2) + (n+1)

but this is just n(n+1) (there are n terms in parentheses, and each of these n terms is equal to n+1).

Thus S(n) = n(n+1)/2.

4) Now consider how S(n)^2 changes as n -> n+1

S(n+1)^2 = (S(n) + (n+1))^2 = S(n)^2 + 2(n+1)S(n) + (n+1)^2

so

S(n+1)^2 - S(n)^2 = 2(n+1)S(n) + (n+1)^2

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

= n^3 + 2n^2 + n + n^2 + 2n + 1

= n^3 + 3n^2 + 3n +1

= (n+1)^3

and we are done.

The one that makes the bigger number

For the warmup problem, A is clearly larger than B: (a+b)^2 = a^2 + ab + b^2, which is clearly larger than a^2 + b^2 for all positive integers.

Now, the actual problem? By inspection, it does appear to hold true for smaller values.

So, what I did was show that, if you solve the equation (1+n)^2 = 1^3 + n^3, the only positive solution is n = 2. If you then continue on and repeat the process by adding the next positive integer into the pattern and another n, the solution to n is always going to be the next positive integer. Does that count as a proof? Hell if I know.