Using the binomial theorem:
[tex](a+b)^n=\Sigma(\frac{n}{i})a^{n-i}*b^i[/tex]
In this case n=17; a=x and b=3, the sum starts in i=0 until n.
And :
[tex]\begin{bmatrix}{n} & {\placeholder{⬚}} \\ {i} & {\placeholder{⬚}}\end{bmatrix}=\frac{n!}{(n-i)!*i!}[/tex]
Substituing the values:
First term: i=0.
[tex](x+3)^{17}=x^{17-0}*3^0*\frac{17!}{(17-0)!*0!}=x^{17}*\frac{17!}{17!}=x^{17}[/tex]
Remember that 0!=1.
Second term: i=1
[tex]x^{17-1}*3^1*\frac{17!}{(17-1)!*1!}=x^{16}*3*\frac{17*16!}{16!}=x^{16}*3*17=51x^{16}[/tex]
Third term: i=2.
[tex]\begin{gathered} x^{17-2}*3^2*\frac{17!}{(17-2)!*2!}=x^{15}*9*\frac{17*16*15!}{15!*2!} \\ =x^{15}*9*\frac{17*16}{2}=1224x^{15} \end{gathered}[/tex]
Answer:
x^17 + 51*x^16 + 1224*x^15.
