Chapter 10, Problem 12STP

To determine

Use binomial theorem to expand the expression: ${(c+d)}^6$

Answer to Problem 12STP

  ${(c+d)}^6=c^6+6c^{6-1}d+15c^4{d}^2+20c^3{d}^3+15c^2{d}^4+6c{d}^5+d^6$

Explanation of Solution

Given: The expression: ${(c+d)}^6$

Concept Used:

Binomial theorem for expand the expression ${(a+b)}^n={\displaystyle \sum _{k=0}^n{}^n{\text{C}}_k·a^{n-k}·b^k}\Rightarrow {\displaystyle \sum _{k=0}^n\frac{n!}{k!(n-k)!}·a^{n-k}·b^k}$

Calculation:

  ${(a+b)}^n={\displaystyle \sum _{k=0}^n{}^n{\text{C}}_k·a^{n-k}·b^k}\Rightarrow {\displaystyle \sum _{k=0}^n\frac{n!}{k!(n-k)!}·a^{n-k}·b^k}$

  $\begin{array}{}$ {(c+d)}^6={}^6{\text{C}}_0·c^{6-0}·{(d)}^0{+}^6{\text{C}}_1·c^{6-1}·{(d)}^1{+}^6{\text{C}}_2·c^{6-2}·{(d)}^2$$ {+}^6{\text{C}}_3·c^{6-3}·{(d)}^3+{}^6{\text{C}}_4·c^{6-4}·{(d)}^4{+}^6{\text{C}}_5·c^{6-5}·{(d)}^5{+}^6{\text{C}}_6·c^{6-6}·{(d)}^6$\end{array}$

  ${(c+d)}^6=c^6+6c^{6-1}d+15c^4{d}^2+20c^3{d}^3+15c^2{d}^4+6c{d}^5+d^6$

Thus, ${(c+d)}^6=c^6+6c^{6-1}d+15c^4{d}^2+20c^3{d}^3+15c^2{d}^4+6c{d}^5+d^6$