Binomial Coefficients for $(x+1)^4$

Question

Find $(x + 1)^4$ using binomial coefficients.

I'm confused as to how to start this, as I thought binomial coefficients were things like $9 \choose 2$.

Answer 1

HINT: $$\binom nr=\frac{n!}{(n-r)! r!}=\frac{n(n-1)\cdots(n-r+1)}{r!}$$

So, $\binom 92=\frac{9(9-1)}{2!}=\frac{9\cdot8}{2\cdot1}=36$

$$(x+1)^4=\sum_{0\le r\le 4}\binom 4r(1)^{4-r}x^r$$

Answer 2

Use this: $(x+y)^n=$ $n \choose 0 $ $x^n y^0$+$n \choose 1 $ $x^{n-1}y^1$+$n \choose 2 $ $x^{n-2}y^2$+ $\dots$ +$n \choose n $ $x^0y^n$

What's the $r^{th}$ term's coefficients using this?

Answer 3

Use $$(x+y)^n=\sum_{r=0}^n T_r$$ $$T_r= \ ^nC_r\times x^{r}y^{n-r}$$

so for $(1+x)^n $you come with $$T_r= \ ^nC_r x^r$$

$$^nC_r=\dfrac{n!}{(n-r)!r!}$$

Answer 4

Do you know the binomial theorem?

$$(a+b)^n = \binom n0 a^nb^0 + \binom n1 a^{n-1}b^1 + \binom n2 a^{n-2}b^2 +\cdots + \binom nn a^0b^n $$