Chapter 8.4, Problem 70E

To determine

The co-efficient of the term $z^{14}{y}^5$ in the Binomial expansion of ${\left(z^2+y\right)}^{12}$ .

Answer to Problem 70E

The co-efficient the term $z^{14}{y}^5$ is $792$ .

Explanation of Solution

Given information :

Givenexpression is ${\left(z^2+y\right)}^{12}$

Formula used:

  ${\left(r+1\right)}^{th}$ term in the expansion of a binomial ${\left(x+y\right)}^n$ is given by $C{}_r{x}^{n-r}{y}^r$

And the coefficient of

  $x^{n-r}{y}^r$ is ${}_n{C}_r=\frac{n!}{\left(n-r\right)!r!}$

Calculation:

Consider the given binomial.

  ${\left(z^2+y\right)}^{12}$

Let ${\left(r+1\right)}^{th}$ term is of $z^{14}{y}^5$ .

Now, compare the above binomial with ${\left(x+y\right)}^n$ .

  $n=12,x=z^2,$

Now, calculate the ${\left(r+1\right)}^{th}$ term of the expansion as shown:

  $\begin{array}{c}C{}_r{x}^{n-r}{y}^r=C{}_r{\left(z^2\right)}^{12-r}{\left(y\right)}^r\\ =C{}_r{\left(x\right)}^{24-2r}{y}^r\end{array}$

Again, ${\left(z\right)}^{24-2r}{y}^r=z^{14}{y}^5$

Therefore,

  $\begin{array}{c}24-2r=14\\ r=5\end{array}$

Now, the 6^th term is the term of $z^{14}{y}^5$ .

Coefficient of 6^th term can be calculated as shown:

  $\begin{array}{c}C{}_5=\frac{12!}{7!5!}\\ =792\end{array}$

Hence, the coefficient of $z^{14}{y}^5$ in the expansion of given binomial is $792$ .