Chapter 12.6, Problem 2CFU

(a)

To determine

To find: Theidentification sign of the second term for each expansion.

Answer to Problem 2CFU

By observing the above three expansions, the sign of the second term of each expansion is negative.

Explanation of Solution

Given:

  ${\left(x-y\right)}^n$ for $n=3,4\text{ and }5.$

Concept used:

Binomial expansion:

  ${\left(x+y\right)}^n={\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{y}^r}.$

  ${\left(x-y\right)}^n={\left(x+\left(-y\right)\right)}^n={\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{\left(y\right)}^r{\left(-1\right)}^r}$ .

Calculation:

First, find the expansion of ${\left(x-y\right)}^n$ .

  ${\left(x+y\right)}^n={\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{y}^r}.$

Substitute $-y$ in replaced of $y$ in the above equation, it gets:

  $\begin{array}{l}{\left(x-y\right)}^n={\left(x+\left(-y\right)\right)}^n.\\ \Rightarrow {\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{\left(-y\right)}^r}.\\ \Rightarrow {\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{\left(y\right)}^r{\left(-1\right)}^r}.\end{array}$

By using the above formula:

  ${\left(x-y\right)}^3={\displaystyle \sum _{r=0}^3\frac{3!}{r!\left(n-r\right)!}{x}^{3-r}{\left(y\right)}^r{\left(-1\right)}^r}.$

By putting the value from the given:

  $\frac{3!}{0!\left(3-0\right)!}=1,\frac{3!}{1!\left(3-1\right)!}=3,\frac{3!}{2!\left(3-2\right)!}=3\text{ and }\frac{3!}{3!\left(3-3\right)!}=1.$

Therefore, the equation can be written as:

  $\begin{array}{l}{\left(x-y\right)}^3=1.x^{3-0}{\left(-y\right)}^0+3.x^{3-1}{\left(-y\right)}^1+3.x^{3-2}{\left(-y\right)}^2+1.x^{3-3}{\left(-y\right)}^3.\\ \Rightarrow x^3-3x^{2}y+3x{y}^2-y^3.\end{array}$

In this expansion, first and third terms are positive but second and fourth terms are negative.

Similarly,

For

  $\begin{array}{l}n=4;\\ {\left(x-y\right)}^4=x^4-4x^{3}y+6x^2{y}^2-4x{y}^3+y^4.\\ n=5;\\ {\left(x-y\right)}^5=x^5-5x^{4}y+10x^3{y}^2-10x^2{y}^3+5x{y}^4-y^5.\end{array}$

Hence, by observing the above three expansions, the sign of the second term of each expansion is negative.

(b)

To determine

To find:the identification the sign of the third term for the expansion.

Answer to Problem 2CFU

By observing the above three expansions, the sign of the third term of each expansion is positive.

Explanation of Solution

Given:

  ${\left(x-y\right)}^n$ for $n=3,4\text{ and }5.$

Concept used:

Binomial expansion:

  ${\left(x+y\right)}^n={\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{y}^r}.$

  ${\left(x-y\right)}^n={\left(x+\left(-y\right)\right)}^n={\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{\left(y\right)}^r{\left(-1\right)}^r}$ .

Calculation:

First, find the expansion of ${\left(x-y\right)}^n$ .

  ${\left(x+y\right)}^n={\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{y}^r}.$

Substitute $-y$ in replaced of $y$ in the above equation, it gets:

  $\begin{array}{l}{\left(x-y\right)}^n={\left(x+\left(-y\right)\right)}^n.\\ \Rightarrow {\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{\left(-y\right)}^r}.\\ \Rightarrow {\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{\left(y\right)}^r{\left(-1\right)}^r}.\end{array}$

By using the above formula:

  ${\left(x-y\right)}^3={\displaystyle \sum _{r=0}^3\frac{3!}{r!\left(n-r\right)!}{x}^{3-r}{\left(y\right)}^r{\left(-1\right)}^r}.$

By putting the value from the given:

  $\frac{3!}{0!\left(3-0\right)!}=1,\frac{3!}{1!\left(3-1\right)!}=3,\frac{3!}{2!\left(3-2\right)!}=3\text{ and }\frac{3!}{3!\left(3-3\right)!}=1.$

Therefore, the equation can be written as:

  $\begin{array}{l}{\left(x-y\right)}^3=1.x^{3-0}{\left(-y\right)}^0+3.x^{3-1}{\left(-y\right)}^1+3.x^{3-2}{\left(-y\right)}^2+1.x^{3-3}{\left(-y\right)}^3.\\ \Rightarrow x^3-3x^{2}y+3x{y}^2-y^3.\end{array}$

In this expansion, first and third terms are positive but second and fourth terms are negative.

Similarly,

For

  $\begin{array}{l}n=4;\\ {\left(x-y\right)}^4=x^4-4x^{3}y+6x^2{y}^2-4x{y}^3+y^4.\\ n=5;\\ {\left(x-y\right)}^5=x^5-5x^{4}y+10x^3{y}^2-10x^2{y}^3+5x{y}^4-y^5.\end{array}$

Hence, by observing the above three expansions, the sign of the third term of each expansion is positive.

(c)

To determine

To find:the explanation to determine the sign of a term without writing out the entire expansion.

Answer to Problem 2CFU

Using the mentioned statements, it can decide the sign of the term in the expansions.

Explanation of Solution

Given:

  ${\left(x-y\right)}^n$ for $n=3,4\text{ and }5.$

Concept used:

Binomial expansion:

  ${\left(x+y\right)}^n={\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{y}^r}.$

  ${\left(x-y\right)}^n={\left(x+\left(-y\right)\right)}^n={\displaystyle \sum _{r=0}^n\frac{n!}{r!\left(n-r\right)!}{x}^{n-r}{\left(y\right)}^r{\left(-1\right)}^r}$ .

Calculation:

Here it can be clearly observed that the sign of first term is positive and the sign of the second term is negative in the three binomials.

  ${\left(x-y\right)}^3,{\left(x-y\right)}^4,{\left(x-y\right)}^5.$

Particularly, the sign of the terms in the expansion depends upon the exponent of second term only because, second term is negative and the first term is positive.

If the exponent of second term is an even number, then the sign is positive.

And if the exponent of the second term is an odd number, then the sign is negative.

But an exponent of first term is either positive or negative, the sign is positive.

Hence, by using thementioned statements, it can decide the sign of the term in the expansions.