How do you use the binomial series to expand # (x + 3)^12 #?
1 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
Write out the row starting
#1# ,#12# ,#66# ,#220# ,#495# ,#792# ,#924# ,#792# ,#495# ,#220# ,#66# ,#12# ,#1#
Write out ascending powers of
#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#
