How do you use the binomial series to expand #(2+1/4x)^9#?

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Answer 1 · 2016-11-17T16:49:24

#512( 1 + 9(x/8)+36(x/8)^2 + 84(x/6)^3 + 126(x/8)^4#

#+ 126(x/6)^5 + 84(x/8)^6 + 36(x/8)^7 + + 9(x/8)^8 + (x/8)^9)#

The binomial series for #( a + b )^N, N= 2. 3. 4, ...# is

#a^N+NC_1a^(N-1)b+NC_2a^(N-2)b^2+...+NC_(N-1)ab^(N-1)+b^N#

where, #NC_r = 1/(r!)N(N-1)(N-2)...(N-r)-NC_(N-r)#.

Using #9C_r=9C_(9-r)#, r = 1, 2, 3 and 4,

#(2+1/4x)^9#

#=2^9(1+x/8)^9# ( to make a = 1 )

#= 512( 1 + 9(x/8)+36(x/8)^2 + 84(x/8)^3 + 126(x/8)^4#

#+ 126(x/8)^5 + 84(x/8)^6 + 36(x/8)^7 + + 9(x/8)^8 + (x/8)^9)#