Chapter 6.1, Problem 36E

Solution

To calculate: The sum of the numbers for each row of 0-4 and the $\text{n}$ th row.

1, 2, 4, 8, and ${\text{2}}^{n-1}$

Given information:

The Pascal Triangle is given as

  

Calculation:

Every numbers in Pascal’s Triangle except 1 is some of the closest two numbers in row directly above it.

Pascal’s Triangles for combination is shown below.

  

Pascal’s Triangles as numbers can be written as

  $\begin{array}{ccccccccccccccc}& & & & & & & 1& & & & & & & \\ & & & & & & 1& & 1& & & & & & \\ & & & & & 1& & 2& & 1& & & & & \\ & & & & 1& & 3& & 3& & 1& & & & \\ & & & 1& & 4& & 6& & 4& & 1& & & \\ & & 1& & 5& & 10& & 10& & 5& & 1& & \\ & 1& & 6& & 15& & 20& & 15& & 6& & 1& \\ 1& & 7& & 21& & 105& & 105& & 21& & 7& & 1\end{array}$

Sum of all elements in first row is 1, second row is 2, 3^rd row is 4, 4^th row is 8, 5^th row is 16.

The pattern continues.

For $\text{n}=2$ , sum of all elements of that row is

  $\begin{array}{c}{\text{2}}^{n-1}=2^{2-1}\\ =2\end{array}$

For $\text{n}=3$ , sum of all elements of that row is

  $\begin{array}{c}{\text{2}}^{3-1}=2^{3-1}\\ =2^2\\ =4\end{array}$

For $\text{n}=4$ , sum of all elements of that row is

  $\begin{array}{c}{\text{2}}^{4-1}=2^{4-1}\\ =2^3\\ =8\end{array}$

For $\text{n}=5$ , sum of all elements of that row is

  $\begin{array}{c}{\text{2}}^{5-1}=2^{5-1}\\ =2^4\\ =16\end{array}$

So, sum of all elements of each row is twice the previous row and 2 raised to the power of row number decreased by 1.

So, the sum of all elements of $\text{n}$ th row will be ${\text{2}}^{n-1}$ .