## Content

### Further identities and results

#### Sum of the squares of the binomial coefficients

In the following diagram of Pascal's triangle, we see that by summing the squares of the

numbers in the fourth row,

\[ 1^2+ 4^2+6^2+4^2+1^2=70, \]we get the middle number of the eighth row.

Detailed description of diagram

We now explain the general pattern.

Using the fact that

\[ \dbinom{n}{r} = \dbinom{n}{n-r}, \]we can write the binomial expansion in two different ways:

\begin{align*} (1+x)^n &= \dbinom{n}{0}+ \dbinom{n}{1}x + \dbinom{n}{2}x^2+\dots+\dbinom{n}{r}x^r+\dots+ \dbinom{n}{n-1}x^{n-1} + \dbinom{n}{n}x^n,\\ (1+x)^n &= \dbinom{n}{n}+ \dbinom{n}{n-1}x + \dbinom{n}{n-2}x^2+\dots+\dbinom{n}{r}x^{n-r}+\dots+ \dbinom{n}{1}x^{n-1} + \dbinom{n}{0}x^n. \end{align*}Multiplying these two expansions together, we see that in \((1+x)^n(1+x)^n\) the

coefficient of \(x^n\) is

\[ \dbinom{n}{0}^2 + \dbinom{n}{1}^2+\dots+\dbinom{n}{r}^2+\dots+\dbinom{n}{n}^2. \]Also, since

\[ (1+x)^n (1+x)^n = (1+x)^{2n}, \]the coefficient of \(x^n\) in this expansion is \(\dbinom{2n}{n}\). Equating the two expressions

for the coefficient of \(x^n\) we have

\[ \dbinom{2n}{n}= \dbinom{n}{0}^2 + \dbinom{n}{1}^2+\dots+ \dbinom{n}{r}^2+\dots+\dbinom{n}{n}^2. \]Using summation notation, this identity is

\[ \dbinom{2n}{n}=\sum_{r=0}^{n}\dbinom{n}{r}^2. \]The case for \(n = 4\) was given at the start of this section:

\[ \dbinom{8}{4} = \dbinom{4}{0}^2 + \dbinom{4}{1}^2+ \dbinom{4}{2}^2 + \dbinom{4}{3}^2+ \dbinom{4}{4}^2. \]##### Exercise 10

The number of forward paths from the bottom left to the top right of an \(n\times n\) grid

is \(\dbinom{2n}{n}\). Use an argument involving paths in the grid to prove

\[ \dbinom{2n}{n}= \dbinom{n}{0}^2 + \dbinom{n}{1}^2+\dots+ \dbinom{n}{r}^2+\dots+\dbinom{n}{n}^2. \](Hint. Draw a diagonal line across the grid from the top-left corner to the bottom-right corner. Choose a point of the grid on the diagonal line. Consider the number of forward paths from the bottom-left corner to the point and then the number of paths from the point to the top-right corner. Repeat for each point on the diagonal line.)

#### Yet more results and identities

We give some further results that come from applying the binomial theorem to \((1 + x)^n\).

##### Exercise 11

Each of the following expansions has three coefficients in arithmetic progression:

\begin{align*} (1+x)^7 &= 1+7x+21x^2+35x^3+35x^4+21x^5+7x^6+x^7,\\ (1+x)^{14} &= 1+14x+91x^2+364x^3+1001x^4+2002x^5+3003x^6+3432x^7+\dotsb,\\ (1+x)^{23} &= \dotsb + 490\,314 x^8 + 817\,190 x^9 + 1\,144\,066 x^{10} + \dotsb. \end{align*}In the first expansion, where \(n=7\), the arithmetic progression is 7, 21, 35.

In the second expansion, where \(n=14\), the progression is 1001, 2002, 3003.

For the expansion of \((1+x)^n\), where \(n>2\), prove that if the coefficients of three

consecutive powers are in arithmetic progression, then \(n+2\) is a perfect square.

(Hint. Consider three coefficients \(\dbinom{n}{r-1}\), \(\dbinom{n}{r}\), \(\dbinom{n}{r+1}\).)

#### Example

Prove the identity

\[ n\times2^{n-1}= \sum_{k=0}^{n}k\times\dbinom{n}{k} \]and check in the case when \(n=4\).

#### Solution

We begin with the expansion of \((1 + x)^n\) from the binomial theorem:

\[ (1+x)^n= \sum_{k=0}^{n}\dbinom{n}{k}x^k. \]Differentiating both sides of this identity with respect to \(x\) gives

\[ n(1+x)^{n-1}= \sum_{k=0}^{n}k\dbinom{n}{k}x^{k-1}. \]Now substitute \(x=1\) to obtain the result:

\[ n2^{n-1}= \sum_{k=0}^{n}k\dbinom{n}{k}. \]When \(n=4\), the left-hand side is \(4\times2^3 = 32\) and the right-hand side is

\begin{align*} 4+2\times\dbinom{4}{2} +3\times\dbinom{4}{3}+4 &=4+2\times6+3\times4+4\\ &=32. \end{align*}If we substitute \(x = -1\) into the identity from the previous example,

\[ n(1+x)^{n-1}= \sum_{k=0}^{n}k\dbinom{n}{k}x^{k-1}, \]then we obtain

\[ \sum_{k=0}^{n}k\dbinom{n}{k}(-1)^{k-1} = 0. \]For example, when \(n=6\) this gives

\[ \dbinom{6}{1} +3 \dbinom{6}{3} + 5\dbinom{6}{5} = 2\dbinom{6}{2} + 4\dbinom{6}{4}+ 6\dbinom{6}{6}. \]#### A link to the Fibonacci numbers

Let \(F(n)\) denote the sum

\[ F(n) = \dbinom{n}{0} + \dbinom{n-1}{1} + \dbinom{n-2}{2} + \dotsb \]where we keep summing until the lower number exceeds the top one. Then

\[ F(n+1) = \dbinom{n+1}{0} + \dbinom{n}{1} + \dbinom{n-1}{2} + \dotsb. \]Adding the first term in line 1 with the second in line 2 and so on, we get

\[ F(n+1) + F(n) = \dbinom{n+1}{0} + \Bigg[\dbinom{n}{0} + \dbinom{n}{1}\Bigg] + \Bigg[\dbinom{n-1}{1} + \dbinom{n-1}{2}\Bigg] + \dotsb. \]Using Pascal's identity, we obtain

\[ F(n+1)+F(n) = \dbinom{n+2}{0}+ \dbinom{n+1}{1}+ \dbinom{n}{2} +\cdots = F(n+2). \]Since \(F(0)=F(1)=1\), we have shown that \(F(0)\), \(F(1)\), \(F(2)\), \(F(3)\), \(\dots\) is the sequence of **Fibonacci numbers**.

This result can be demonstrated in terms of Pascal's triangle.