How do you find the first 3 non-zero terms for the expansion of #(9+x^2)^ (1/2)#?

1 Answer
Apr 29, 2017

Answer:

#(9+x^2)^(1/2)=3+x^2/6-x^4/216+x^6/3888...............#

Explanation:

Binomial expansion of #(1+x)^n=C_0^na^n+C_1^nx+C_2^nx^2+C_3^nx^3+..++C_n^nx^n#,

where #C_r^n=(n(n-1)(n-2).....(n-r+1))/(1.2.3.....r)# and #C_0^n=1#

When we have #n# as fraction this is written as infinite series.

Hence #(9+x^2)^(1/2)# can be written as

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

= #3[1+(1/2)/1(x^2/9)+(1/2(1/2-1))/(1*2)(x^2/9)^2+(1/2(1/2-1)(1/2-2))/(1*2*3)(x^2/9)^3+............]#

= #3[1+x^2/18+(1/2*(-1/2))/2(x^4)/81+(1/2*(-1/2)(-3/2))/6(x^6)/729.........]#

= #3[1+x^2/18-x^4/648+x^6/11664...............]#

= #3+x^2/6-x^4/216+x^6/3888...............#