# Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on

I am noticing this pattern: \begin{align} 1+2&=3\\ 4+5+6&=7+8\\ 9+10+11+12&=13+14+15 \\ 16+17+18+19+20&=21+22+23+24 \\ &\vdots \end{align} Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire summation? I think the $n$th term starts with $$n^2+(n^2+1)+\cdots=\cdots+[(n+1)^2-1].$$ I am also presuming once this formula is discovered, we can prove it by induction for any $n$.

• The other interesting thing here is that 1,2,3, etc. appear in order in the list. And you have 2,3,4, etc. terms on the left, 1,2,3, etc. terms on the right. This should let you determine a formula like the one you want. Then prove it by induction. – GEdgar Jan 7 '15 at 19:08
• Adding $n^2$ to $n$ terms is the same as adding $n$ to each term. – user88524 Jan 7 '15 at 19:29
• Your observation that the $n$th line starts with $n^2$ is a good one, but it's easy to see that it doesn't, in general, end in $(n^2+3)$. It must instead end in $(n+1)^2-1$, in order for the next line to start with $(n+1)^2$. – Barry Cipra Jan 7 '15 at 19:52

## 7 Answers

The formula would appear to be:

$n^2+(n^2+1)+\cdots+(n^2+n)=(n^2+n+1)+\cdots+(n^2+2n)$

Note that there are (n+1) terms on the left hand side and n terms on the right. As for a closed form of the sum, note that each side is an arithmetic series with a common difference of 1 and there are formulas you could use to compute this.

Note that the $n^2+2n+1=(n+1)^2$ for how the next row starts makes sense in the end as well.

An attempt at the closed form of the expression would be:

$n^2(n+1)+\frac{n(n+1)}{2}$

Where the first term is the $(n+1)$ occurances of $n^2$ whereas the second is the sum of the first $n$ natural numbers which is a well-known formula. If you want the sum of a row, you could just double this sum.

Hopefully a proof without words. • I was looking for a look-see proof when you posted this. – John Jan 7 '15 at 20:02
• @Gamamal I really could not understand it...is there a link I can follow? – MonK May 20 '15 at 18:10
• The left side of the equation is in red and orange, and the right side of the equation is green. He takes the last term of the sum (in orange) and distributes it to the other three (in the bottom row). By rearranging the pieces this way, one can easily see that the left side and the right side are identical and hence add to the same amount. – user242594 Jun 18 '15 at 3:13

Proof of the Identities \begin{align} n^2&=\sum_{k=1}^nn\tag{1}\\ n^2+\sum_{k=1}^n\left(n^2+k\right)&=\sum_{k=1}^n\left(n^2+n+k\right)\tag{2}\\ \sum_{k=0}^n\left(n^2+k\right)&=\sum_{k=n+1}^{2n}\left(n^2+k\right)\tag{3} \end{align} Explanation:
$(1)$: multiplication written as a sum
$(2)$: add $\sum\limits_{k=1}^n\left(n^2+k\right)$ to both sides
$(3)$: include $n^2$ in the sum on the left and reindex the sum on the right

Derivation of the Formula for Each Side

Using the formula $$\sum_{k=1}^nk=\frac{n(n+1)}2\tag{4}$$ and equation $(2)$, we can compute the sum of each side of $(3)$ as \begin{align} n^2+n^3+\frac{n(n+1)}2 &=\frac{2n^3+3n^2+n}2\\ &=\frac{n(n+1)(2n+1)}2\tag{5} \end{align}

Both sides are $n (n+1)(2n+1)/2$ which also happens to be $3 (1^2 + 2^2 + \ldots + n^2)$.

$n^2 = \sum_{k=0}^{n-1} n = \sum_{k=0}^{n-1} ((n(n+2)-k)-(n(n+1)-k))$, since $(n(n+2)-k)-(n(n+1)-k) = n$.

So we have $n^2+\sum_{k=0}^{n-1} (n(n+1)-k) = \sum_{k=0}^{n-1} (n(n+2)-k)$.

Let's examine a simple identity: $$9+10+11+12=13+14+15$$ Notice that there are $4$ terms on the left and $3$ on the right - so let's "use up" one of the terms on the left. We can split $9=3+3+3$ and distribute these additional $+3$ terms to the other $3$ summands on that side to get: $$(10+\color{red} 3)+(11+\color{red} 3)+(12+\color{red} 3)=13+14+15$$ and, of course, the left hand is exactly $13+14+15$ when we compute. Well, that was easy!

Let's generalize. The sum is always of the form $$n^2+(n^2+1)\ldots+(n^2+n)=(n^2+n+1)+\ldots + (n^2+2n)$$ and there are $n+1$ terms on the left hand and $n$ terms on the right hand side. So how do we get from left to right? Easy! We take the $n^2$ term and divide it into $\underbrace{n+n+\ldots+n}_{n\text{ times}}$. Now, there are $n$ other summands on that side, so we distribute a $+n$ term to each of them without changing the value. This yields $$(n^2+1+\color{red} n)+\ldots+(n^2+n+\color{red}n)=(n^2+n+1)+\ldots + (n^2+2n)$$ but this is obvious, since simplifying both sides gives $$(n^2+n+1)+\ldots+(n^2+2n)=(n^2+n+1)+\ldots + (n^2+2n)$$ where both sides are identical.

Another way to build up the pattern you see.

Start with the $n$th triangle number, $T_n$ (I'll use $n=5$):

$$1 + 2 + 3 + 4 + 5 = 1 + 2 + 3 + 4 + 5.$$

Add $n^2$ to both sides, except distribute the $n^2$ on one side by adding $n$ to each of the $n$ terms:

$$5^2 + 1 + 2 + 3 + 4 + 5 = 6 + 7 + 8 + 9 + 10.$$

Finally, add $n^3$ to both sides by distributing $n^2$ to all terms except the first one on the left:

$$25 + 26 + 27 + 28 + 29 + 30 = 31 + 32 + 33 + 34 + 35.$$

So, both sides are

$$T_n + n^2 + n^3 = \frac{n(n+1)}{2} + n^2 + n^3.$$