# How do you use the Binomial Theorem to expand (d+3)^7?

Jul 4, 2016

$= {d}^{7} + 21 {d}^{6} + 189 {d}^{5} + 945 {d}^{4} + 2835 {d}^{3} + 5103 {d}^{2} + 5103 d + 2187$

#### Explanation:

The binomial theorem is given by:

${\left(x + y\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {x}^{n - k} {y}^{k}$

where ((n),(k)) = (n!)/(k!(n-k)!) taking note that 0! = 1

${\left(d + 3\right)}^{7} = {\sum}_{k = 0}^{7} \left(\begin{matrix}7 \\ k\end{matrix}\right) {d}^{7 - k} {3}^{k}$

$= \left(\begin{matrix}7 \\ 0\end{matrix}\right) {d}^{7} {3}^{0} + \left(\begin{matrix}7 \\ 1\end{matrix}\right) {d}^{6} {3}^{1} + \left(\begin{matrix}7 \\ 2\end{matrix}\right) {d}^{5} {3}^{2.} . . \left(\begin{matrix}7 \\ 7\end{matrix}\right) {d}^{0} {3}^{7}$

= (7!)/(0!7!)d^7 + (7!)/(1!6!)d^(6)*3 + (7!)/(2!5!)d^5*9 ... (7!)/(7!0!)*3^7

$= {d}^{7} + 21 {d}^{6} + 189 {d}^{5} + 945 {d}^{4} + 2835 {d}^{3} + 5103 {d}^{2} + 5103 d + 2187$

Note, for brevity I have skipped out a chunk of the calculation but the method is the exact same so you should have no issue with reproducing it