Chapter 9.5, Problem 102E

To determine

To prove: The sum of the numbers in the nth row of Pascal’s Triangle is $2^n$

Explanation of Solution

Concept used:

Binomial theorem:

  ${\left(x+y\right)}^n=C{}_0{x}^n{y}^0+C{}_1{x}^{n-1}y+C{}_2{x}^{n-2}{y}^2+\cdots +C{}_{n-1}x{y}^{n-1}+C{}_n{x}^0{y}^n$

Where $C{}_r=\frac{n\text{!}}{\left(n-r\right)\text{!}r\text{!}}$

Pascal’s triangle:

The n^th row of Pascal’s triangle gives the coefficients of ${\left(x+y\right)}^n$ .

It is known that the n^th row of Pascal’s triangle gives the coefficients of ${\left(x+y\right)}^n$ .

That is, $C{}_r$ represents the r^th element of the n^th row of the Pascal’s triangle for $r=0,1,2,\mathrm{...},n$ .

So, the sum of the numbers in the nth row of Pascal’s Triangle is represented as $C{}_0+C{}_1+C{}_2+\cdots +C{}_n$ .

Now, use binomial theorem to expand ${\left(1+1\right)}^n$

  $\begin{array}{c}{2}^n={\left(1+1\right)}^n\\ =C{}_0{\left(1\right)}^n{\left(1\right)}^0+C{}_1{\left(1\right)}^{n-1}{\left(1\right)}^1+C{}_2{\left(1\right)}^{n-2}{\left(1\right)}^2+\cdots +C{}_n{\left(1\right)}^0{\left(1\right)}^n\\ =C{}_0+C{}_1+C{}_2+\cdots +C{}_n\end{array}$

Hence proved that, the sum of the numbers in the nth row of Pascal’s Triangle is $2^n$ .