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

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?

• Hint: Try grouping the even terms and taking someting common out. Commented Jun 27, 2021 at 11:35

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 $$8C = B$$.