How do you use the binomial series to expand # (x + 3)^9 #?

1 Answer
Jun 19, 2017

Answer:

#:.(x+3)^9 =x^9+27x^8+324x^7 + 2268 x^6 + 10206 x^5+30618x^4+61236x^3+78732x^2+59049x+19683#

Explanation:

We know #(a+b)^n= nC_0 a^n*b^0 +nC_1 a^(n-1)*b^1 + nC_2 a^(n-2)*b^2+..........+nC_n a^(n-n)*b^n#

Here #a=x,b=3,n=9# We know, #nC_r = (n!)/(r!*(n-r)!#
#:.9C_0 =1 ,9C_1 =9, 9C_2 =36,9C_3 =84, 9C_4 =126,9C_5 =126,9C_6 =84, 9C_7 =36 ,9C_8 =9, 9C_9 =1 #

#:.(x+3)^9 = x^9+9*x^8*3+36*x^7*3^2+84*x^6*3^3+126*x^5*3^4+126*x^4*3^5+84*x^3*3^6+36*x^2*3^7+9x*3^8+3^9# or

#:.(x+3)^9 =x^9+27x^8+324x^7 + 2268 x^6 + 10206 x^5+30618x^4+61236x^3+78732x^2+59049x+19683# [Ans]