Question

X+1 ^ 16 what is the coefficient of x^5

Answer 1

Binomial theorem:

$(x+1)^{16}=\displaystyle\sum_{k=0}^{16}\dbinom{16}kx^{16-k}1^k=\sum_{k=0}^{16}\dbinom{16}kx^{16-k}$

The $x^5$ term corresponds with the term in the sum for which $16-k=5\implies k=11$. For this $k$, we get

$\dbinom{16}{11}x^{16-11}=\dbinom{16}{11}x^5$

So the coefficient you want to find is

$\dbinom{16}{11}=\dfrac{16!}{11!(16-11)!}=4368$