Use a combinatorial proof to show that $1^2 + 2^2 + ... + n^2 = {n+1 \choose 2} + 2{n+1 \choose 3}$

Question

For this question, I tried to list out n+1 objects and count the number of ways by conditioning on whether certain objects were selected or not but that didn't get me anywhere, then I tried to use the method I was taught in class where you are essentially "forming a committee" and tried to count the number of ways I could form a 2 member or 3 member committee out of n+1 members but that didn't work either. I could solve this problem algebraically by transforming the right-hand side to $\frac{n(n+1)(2n+1)}{6}$ and then proving this by induction but the question requires me to give a combinatorial proof and I can't seem to figure out how...

Any help would be greatly appreciated!

Answer 1

I included a poorly drawn picture to make myself a bit more clear. So we have $n$ squares, the first contains a single cell, the next then $2^2$ and so on till $n^2$ cells. If we add the cells in each square separately we get the number $1+2^2+3^2+\ldots+n^2$. If we change the way we count to adding the cells at each level we get:

1. $1+2+\ldots+n$ on the first level
2. $2+3+\ldots+n$ on the second level
3. $3+4+\ldots+n$ on the third level

and so on, I hope you get the idea. The number on the first level is exactly $1+2+\ldots+n=\binom{n+1}{2}$ and on the $i$-th level it is $\binom{n+1}{2}-\binom{i}{2}$. Adding this up gives

\begin{align} \binom{n+1}{2}+\sum\limits_{i=2}^n(\binom{n+1}{2}-\binom{i}{2})&=\binom{n+1}{2}+(n-1)\binom{n+1}{2}-\sum\limits_{i=2}^n\binom{i}{2}\\ &=\binom{n+1}{2}+(n-1)\binom{n+1}{2}-\binom{n+1}{3}\\ &=\binom{n+1}{2}+3\binom{n+1}{3}-\binom{n+1}{3}\\ &=\binom{n+1}{2}+2\binom{n+1}{3}. \end{align}

Don't know whether it is what you were looking for tho, as it uses facts about binomial coefficients.

Answer 2

Given identity: $$ 1^2 + 2^2 + \cdots + n^2 = \binom{n+1}{2} + 2\binom{n+1}{3}, $$

Left-Hand Side: $1^2 + 2^2 + \cdots + n^2$

Consider a group of $n+1$ people labeled $P_1, P_2, \dots, P_{n+1}$. The expression $k^2$ can be interpreted as selecting two ordered people from the first $k$ people (including the case where the two selected people are the same).

Thus, $1^2 + 2^2 + \cdots + n^2$ counts the total number of ways to select ordered pairs $(i, j)$ where both $i$ and $j$ are chosen from the first $k$ people, for each $k = 1, 2, \dots, n$.

Right-Hand Side: $\binom{n+1}{2} + 2\binom{n+1}{3}$

• $\binom{n+1}{2}$ counts the number of distinct unordered pairs from a group of $n+1$ people.
• $2\binom{n+1}{3}$ counts the number of ways to choose three distinct people from the group of $n+1$, and from each group of three, we can form two distinct ordered pairs.

Combinatorial Argument:

Both sides count the same thing: the total number of ways to select ordered pairs of people from a group of $n+1$. The left-hand side counts ordered pairs by summing the possible pairs from different group sizes, while the right-hand side does the same by breaking it into two cases: unordered pairs and pairs chosen from triples of people.

Thus, the equation holds: $$ 1^2 + 2^2 + \cdots + n^2 = \binom{n+1}{2} + 2\binom{n+1}{3}. $$

Verification using algebric method:

The right-hand side can be expanded as: $$ \binom{n+1}{2} = \frac{(n+1)n}{2}, $$ $$ \quad 2\binom{n+1}{3} = 2 \cdot \frac{(n+1)n(n-1)}{6} = \frac{(n+1)n(n-1)}{3}. $$ Thus, $$ \binom{n+1}{2} + 2\binom{n+1}{3} = \frac{(n+1)n}{2} + \frac{(n+1)n(n-1)}{3}. $$ Combining the terms over a common denominator: $$ \frac{(n+1)n}{2} + \frac{(n+1)n(n-1)}{3} = \frac{(n+1)n \left(3 + (n-1)\right)}{6} = \frac{(n+1)n(n+2)}{6}. $$ This is the known formula for the sum of squares: $$ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}. $$ Therefore, the identity is correct.