Finding the sum of the cubes of the first n odd natural numbers.

Question

I know that the sum of the cubes of the first $n$ natural numbers is $({\frac{n(n+1)}{2}})^2$, which in expanded form is $(1)^3+(2)^3+(3)^3+...+(n)^3$. Multiplying by $2^3$ on both sides gives $2(n(n+1))^2$, which I can see is the sum of cube of first $n$ even natural numbers. But my question is why do we take the difference between the sum of cubes of all $2n$ (i.e. $({\frac{2n(2n+1)}{2}})^2$) numbers and sum of cubes of first $n$ even numbers (i.e. $2n^2(n+1)^2$) to obtain the odd sum? How would you distinguish between the two?

Answer 1

Use a specific example.

Let $A = 1^3 + 3^3 + 5^3 + 7^3 + 9^3.$

Let $B = 2^3 + 4^3 + 6^3 + 8^3.$

Let $C = 1^3 + 2^3 + 3^3 + 4^3.$

Assume that you want to evaluate $A$.

You can use the formula to evaluate $(A + B)$, so the problem is reduced to evaluating $B$.

This can be done by using the formula to evaluate $C$, and then reasoning that $8C = B$.