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
