# Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction?

I've been trying to solve the following problem:

Show that $$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$$ using the identity $$1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$$.

I had the following idea, we write:

$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$

Let's pretend the identity we want to prove is true, then:

$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$

We take then $$k-x=j$$ and solve for $$x$$. If the given identities are true, $$x$$ must be the sum of $$2^2+4^2+\dots+(2n)^2$$, and we have that

$$x= \frac{2n (n+1) (2 n+1)}{3}$$

We still don't know that $$x=2^2+4^2+\dots+(2n)^2$$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?

Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.

• Hint: $2^2 + 4^2 + \cdots + (2n)^2 = 4(1^2 + 2^2 + \cdots + n^2 )$
– Gary
Nov 16 '20 at 12:06
• If you take $2^2$ out, you get $1^2 + 2^2 + ...+n^2$ Nov 16 '20 at 12:07
• Hint: Factor out $2^2$ from $x$ so that $x=4(1^2+2^2+\cdots+n^2)$. Then, use the first result. Nov 16 '20 at 12:08
• You don't need induction to know what $x$ is since $x=\sum_n (2n)^2=4\cdot\sum_n n^2$ and for this, you were given the formula. Nov 16 '20 at 12:09
• @quantenstau Damn! I almost noticed that yesterday, I noticed that the sum of the first $n$ powers of even numbers is $4\sum_{i=0}^n (n-i)(n+1)$ but was unable to use this in a useful way. Nov 16 '20 at 12:19

Hint: notice that $$\sum(2k)^2=\sum4k^2=4\sum k^2.$$

Let $$S=1^1+3^2+\cdots+(2n-1)^2=\sum_{k=1}^{n}(2k-1)^2=\sum_{k=1}^{n}[4 k^2-4k+1]$$ $$S=4\sum_{k=1}^n k^2- 4\sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$$ $$S=4\frac{n(n+1)(2n+1)}{6}-4 \frac{n(n+1)}{2}+n=\frac{n(4n^2-1)}{3}$$

Consider the identity $$(2k+1)^2=4k^2+4k+1$$ which relates the sum of the odd squares to the sum of the squares.

The extra terms $$4k+1$$ are easily dealt with with a telescoping,

$$(k+1)^2-k^2=2k+1.$$

Altogether,

$$1^2+3^2+ \dots +(2n+1)^2=4\frac{n(n+1)(2n+1)}6+2(n+1)^2-1-n.$$

$$1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$$ multiplying this by $$2^2$$ gives the sum $$2^2+4^2+\dots + (2n)^2 = \frac{2n(n+1)(2n+1)}{3}$$ and subtracting this from the given identity results int the required equality.

You have a typo. Call the known sum $$P_n$$. The desired sum is\begin{align}P_{2n+1}-4P_n&=\frac{(2n+1)(2n+2)(4n+3)-4n(n+1)(2n+1)}{6}\\&=\frac{(n+1)(2n+1)}{3}(4n+3-2n)\\&=\frac{(n+1)(2n+1)(2n+3)}{3}.\end{align}

Let $S=1^2+3^2+5^2+\cdots+(2n+1)^2$\$.

Next, consider the sums $$S_1=2^2+4^2+6^2+\cdots+(2n)^2$$ $$S_2=1^2+2^2+3^2+\cdots+(2n+1)^2$$ Thus, we can calculate $$S$$ as $$S=S_2-S_1$$.

Let's evaluate $$S_1$$ and $$S_2$$ using the identity we are given.

Clearly, $$S_2=\frac{1}{6}(2n+1)(2n+2)(4n+3)$$ and $$S_1=2^2(1^2+2^2+\cdots+n^2)$$ $$S_1=2^2\cdot\frac{1}{6}n(n+1)(2n+1)$$

Therefore, $$S=S_2-S_1$$ $$S=\frac{1}{6}\left[2(n+1)(2n+1)(4n+3)-4n(n+1)(2n+1)\right]$$ $$S=\frac{(n+1)(2n+1)}{6}\left[8n+6-4n\right]$$ $$S=\frac{(n+1)(2n+1)(2n+3)}{3}$$ as desired.

Hope this helps :)