Respuesta :

yas
binomial theorem

ok so

for (a+b)^n
the first term is a^n
2nd term has a^(n-1)
x^7 is 1 less than x^8
we know that it is the 2nd term

to find the nth term of a bionial expansion (a+b)^n
you do,
the nth term is k, so then you do
k-1=z
then
the coeficint of that is
[tex] \left[\begin{array}{ccc}n\\z\end{array}\right] = \frac{n!}{z!(n-z)!} [/tex]

so n=8
k=2
2-1=1
z=1

[tex] \left[\begin{array}{ccc}8\\1\end{array}\right] = \frac{8!}{1!(8-1)!} [/tex]=[tex] \frac{8!}{(1)(7)!} = \frac{8*7*6*5*4*3*2*1}{7*6*5*4*3*2*1}=8 [/tex]
the coefient is 8