Chapter 4.2, Problem 48E

To determine

To expand: The Binomial ${\left(np-1\right)}^4$ by using the Pascal Triangle.

Answer to Problem 48E

The expansion of the expression ${\left(np-1\right)}^4$ is $n^4{p}^4-4n^3{p}^3+6n^2{p}^2-4np+1$ .

Explanation of Solution

Given information: The expression is ${\left(np-1\right)}^4$ .

Theorem used: The Binomial Theorem: ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\cdots {+}_n{C}_r{x}^{n-r}{y}^r+\cdots +nx{y}^{n-1}+y^n$ , the coefficient of $x^{n-r}{y}^r$ is ${}_n{C}_r=\frac{n!}{\left(n-r\right)!r!}$ .

Calculation:

Here, $n=4$ .

First write out the Pascal’s triangle such that the row that begins with 1, 4.

  $\begin{array}{ccccccccccc}& & & & & 1& & & & & \\ & & & & 1& & 1& & & & \\ & & & 1& & 2& & 1& & & \\ & & 1& & 3& & 3& & 1& & \\ & 1& & 4& & 6& & 4& & 1& \\ & & & & & & & & & & \end{array}$

From the Pascal’s triangle, it is observed that the binomial coefficients of the expression are 1, 4, 6, 4, and 1.

Expand the expression

  ${\left(np-1\right)}^4$ by using formula in Binomial theorem as follows.

  $\begin{array}{c}{\left(np-1\right)}^4=\left(1\right){\left(np\right)}^4{\left(-1\right)}^0+4{\left(np\right)}^{4-1}\left(-1\right)+6{\left(np\right)}^{4-2}{\left(-1\right)}^2+4{\left(np\right)}^{4-3}{\left(-1\right)}^3+1{\left(np\right)}^{4-4}{\left(-1\right)}^4\\ =n^4{p}^4-4{\left(np\right)}^3+6{\left(np\right)}^2-4\left(np\right)+1\\ =n^4{p}^4-4n^3{p}^3+6n^2{p}^2-4np+1\end{array}$

Thus, the expansion of the expression ${\left(np-1\right)}^4$ is $n^4{p}^4-4n^3{p}^3+6n^2{p}^2-4np+1$ .