Chapter 10.5, Problem 56E

To determine

To calculate:

The number of terms for the following:

  ${\left(2s^4+5\right)}^5$

Answer to Problem 56E

The binomial expansion is $32s^{20}+400s^{16}+2000s^{12}+5000s^8+6250s^4+3125$ .

Explanation of Solution

Given information:

  ${\left(2s^4+5\right)}^5$

Formula used:

The binomial theorem:

  $\begin{array}{l}{\left(a+b\right)}^n=a^n+n{a}^{n-1}b+\frac{n\left(n-1\right)}{2!}{a}^{n-2}{b}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{a}^{n-3}{b}^3+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{a}^{n-4}{b}^4+\mathrm{...}+b^n\\ {\left(a+b\right)}^n=C_0{a}^n{b}^0+C_1{a}^{n-1}{b}^1+C_2{a}^{n-2}{b}^2+C_3{a}^{n-3}{b}^3+C_4{a}^{n-4}{b}^4+\mathrm{...}+C_n{a}^0{b}^n\end{array}$

Calculation:

The given expansion is ${\left(2s^4+5\right)}^5$ compared with ${\left(a+b\right)}^n$ , get that: $a=2s^4,b=5,n=5$

The binomial theorem for the given expansion is derived as:

  ${\left(2s^4+5\right)}^5={\left(2s^4\right)}^5+\left(5\right){\left(2s^4\right)}^{5-1}\left(5\right)+\frac{5\left(5-1\right)}{2!}{\left(2s^4\right)}^{5-2}{\left(5\right)}^2+\frac{5\left(5-1\right)\left(5-2\right)}{3!}{\left(2s^4\right)}^{5-3}{\left(5\right)}^3$

  $+\frac{5\left(5-1\right)\left(5-2\right)\left(5-3\right)}{4!}{\left(2s^4\right)}^{5-4}{\left(5\right)}^4+\frac{5\left(5-1\right)\left(5-2\right)\left(5-3\right)\left(5-4\right)}{5!}{\left(2s^4\right)}^{5-5}{\left(5\right)}^5$

  ${\left(2s^4+5\right)}^5={\left(2s^4\right)}^5+\left(5\right){\left(2s^4\right)}^4\left(5\right)+\frac{5\left(4\right)}{2!}{\left(2s^4\right)}^3{\left(5\right)}^2+\frac{5\left(4\right)\left(3\right)}{3!}{\left(2s^4\right)}^2{\left(5\right)}^3$

  $+\frac{5\left(4\right)\left(3\right)\left(2\right)}{4!}{\left(2s^4\right)}^1{\left(5\right)}^4+\frac{5\left(4\right)\left(3\right)\left(2\right)\left(1\right)}{5!}{\left(5\right)}^5\left[\because a^0=1\right]$

  ${\left(2s^4+5\right)}^5=32s^{20}+400s^{16}+2000s^{12}+5000s^8+6250s^4+3125$

Hence, the binomial expansion is $32s^{20}+400s^{16}+2000s^{12}+5000s^8+6250s^4+3125$ .