Chapter 4.7, Problem 69PPE

Solution

a.

To calculate:

Numbers in the seventh row.

Numbers in the seventh row is $1,7,21,35,35,21,7$ .

Given information:

  

Calculation:

Formula for finding out the number in Pascal triangle is calculated by,

  ${\left(x+y\right)}^n=a_0{x}^n+a_1{x}^{n-1}y+a_2{x}^{n-2}{y}^2+\mathrm{...}+a_{n}x{y}^{n-1}+a_n{y}^n$

Numbers in the seventh row of the given Pascal triangle is calculated as,

  $\begin{array}{l}{\left(x+1\right)}^7\\ =1x^7+7x^{6}y+21x^5{y}^2+35x^4{y}^3+35x^3{y}^4+21x^2{y}^5+7x{y}^6+1x{y}^7\end{array}$

So, Numbers in the seventh row is $1,7,21,35,35,21,7$ .

b.

To calculate:

Sum of the numbers in the seventh row.

Sum of the numbers in the seventh row is $128$ .

Given information:

  

Calculation:

Sum of the numbers in each term is ,

  $\begin{array}{l}1+1=2\\ 1+2+1=4\\ 1+3+3+1=8\\ 1+4+6+4+1=16\\ 1+5+10+10+5+1=32\end{array}$

This is ${\left(2\right)}^n$ or ${\left(1+1\right)}^n$ where $n$ is the row no.

In case consider individual row, then it is binomial series.

For e.g.

  $\begin{array}{l}{\left(1+1\right)}^n=2C_0{\left(1\right)}^2+2C_1\left(1\right)\left(1\right)+2C_2{\left(1\right)}^2\\ =1+2+1\end{array}$

Similarly,

  $7^{th}$ row is,

  $\begin{array}{l}{\left(6+1\right)}^n=7C_0={\left(1\right)}^n+7C_1{\left(1\right)}^{7-1}\left(1\right)+7C_2{\left(1\right)}^{7-2}{\left(1\right)}^2+7C_3{\left(1\right)}^{7-3}{\left(1\right)}^3+7C_4{\left(1\right)}^{7-4}{\left(1\right)}^4+7C_5{\left(1\right)}^{7-5}{\left(1\right)}^5+\mathrm{...}+7C_7{\left(1\right)}^{7-7}{\left(1\right)}^7\\ =128\end{array}$