How do you use the binomial series to expand # (x + 3)^12 #?

1 Answer
Dec 30, 2015

Answer:

#(x+3)^12#

#=sum_(k=0)^12 ((12),(k)) 3^k x^(12-k)#

#=x^12+36x^11+594x^10+5940x^9+40095x^8+192456x^7+673596x^6+1732104x^5+3247695x^4+4330260x^3+3897234x^2+2125764x+531441#

Explanation:

You can use Pascal's triangle to obtain the values of #((12),(0))#, #((12),(1))#,...,#((12),(12))#

enter image source here

Write out the row starting #1#, #12#,...

#1#, #12#, #66#, #220#, #495#, #792#, #924#, #792#, #495#, #220#, #66#, #12#, #1#

Write out ascending powers of #3# from #3^0# to #3^12#...

#1#, #3#, #9#, #27#, #81#, #243#, #729#, #2187#, #6561#, #19683#, #59049#, #177147#, #531441#

Multiply both sequences together to get:

#1#, #36#, #594#, #5940#, #40095#, #192456#, #673596#, #1732104#, #3247695#, #4330260#, #3897234#, #2125764#, #531441#

These are the coefficients we need since:

#(x+3)^12 = sum_(k=0)^12 ((12),(k)) 3^k x^(12-k)#

So:

#(x+3)^12#

#=x^12+36x^11+594x^10+5940x^9+40095x^8+192456x^7+673596x^6+1732104x^5+3247695x^4+4330260x^3+3897234x^2+2125764x+531441#