Question

Medal !!
(x + 2)4 = x4 + 8x3 +... Use the Binomial Theorem to complete the binomial expansion.

Answer 1

Use the Binomial theorem to solve this question :

\(\sf{(x+y)^n = \sum_{k=0}^ {n} \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k} {b^k} }\)

Answer 2

Plug-in n = 4 , y = 2 and x = x in the equation above.

Answer 3

I don't know how to do these :(

Answer 4

(x+2)\[(x+2)^{4}=\sum_{k-0}^{4}\left(\begin{matrix}n \\ k\end{matrix}\right)a ^{4-k}b ^{k} ?\]

Answer 5

im so lost

Answer 6

Okay no problem : I will expand that :

(x+2)^4 = \(\sum_{k=0}^{4}\left(\begin{matrix}4 \\ k\end{matrix}\right) x^{4-k} 2^k \)

Answer 7

How do I go about this problem

Answer 8

So, (x+2)^4 =
\[\left(\begin{matrix}4\\ 0\end{matrix}\right)x^{4-0} 2^0 + \left(\begin{matrix}4\\1\end{matrix}\right) x^{4-1} 2^1 + \left(\begin{matrix}4\\2\end{matrix}\right)x^{4-2} 2^2 + \left(\begin{matrix}4\\3\end{matrix}\right) x^{4-3} 2^3 + \left(\begin{matrix}4\\4\end{matrix}\right)x^{4-4} 2^4 \]

Answer 9

Can you solve it now?

Answer 10

I feel like I am looking at alien writing... :(

Answer 11

I see the pattern I just don't know how to solve it.

Answer 12

Okay, see : \(\left(\begin{matrix}4\\0\end{matrix}\right)\) = \(\cfrac{4!}{0!(4-0)!}\) =\(\cfrac{4!}{4!} = 1\)

This is how to solve such patterns.

Answer 13

4! = 4 Factorial = 4 * 3 * 2 * 1

or \(\large{n!} = \textbf{n factorial} = n * (n-1) * ... *3 * 2 * 1\)

Answer 14

So, to solve : \(\left(\begin{matrix}n\\k\end{matrix}\right) = \cfrac{n!}{k!(n-k)!}\)

Answer 15

Are you getting it now?