Proving a formula using another formula

Question

These questions are from the book "What is Mathematics":

Prove formula 1: $$1 + 3^2 + \cdots + (2n+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3}$$ formula 2: $$1^3 + 3^3 + \cdots + (2n+1)^3 = (n+1)^2(2n^2+4n+1)$$

Using formulas 4 and 5; formula 4: $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$ formula 5: $$1^3 + 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$

My approach for proving the first one was to do something like a subtraction between formula 4 and 1, substituting the value of $n $ in formula 4 for $2n + 1$. I was left with a formula which I proved by mathematical induction gives me $2^2 + 4^2 + \cdots + (2n)^2$. I am wondering if this is the correct approach before doing the second proof? Is there a better and simpler way of doing this?

Answer 1

I think you're basically on the right track. Notice that

$$2^2 + 4^2 + 6^2 + \cdots + (2n)^2 = 4(1^2 + 2^2 + 3^2 + ... + n^2).$$

That should make things relatively easy.

Answer 2

I would go like this: $(2k+1)^2 = 4k^2+4k+1 \Rightarrow 1+3^2+5^2+\cdots + (2n+1)^2=\displaystyle \sum_{k=0}^n (2k+1)^2=4\displaystyle \sum_{k=0}^n k^2 + 4\displaystyle \sum_{k=0}^n k + (n+1)= 4\cdot\dfrac{n(n+1)(2n+1)}{6}+4\cdot\dfrac{n(n+1)}{2}+(n+1)=...$