Find the first four terms in the binomial expansion of (1+x)^(1/5) and state the range of values of x for which this expansion is valid.Hence approximate (31)^(1/5) correct to four decimal places.

Answer 1

(1+x)^(1/5) = 1^(1/5) + (1/5)/1! x + (1/5)(-4/5)/2! x^2 + (1/5)(-4/5)(-9/5)/3! x^3

= 1 + x/5 - 2x^2/25 + 6x^3/125

Using the series, we have a problem. It only converges when |x| < 1.

I think what you really want to do is find

(x-1)^(1/5), using x=32. Then you have

x^(1/5) - (1/5)/1!x^(4/5) + (1/5)(-4/5)/2!x^(9/5) - (1/5)(-4/5)(-9/5)/3!x^(14/5)
= x^(1/5) - 1/(5 x^4/5) - 2/(25 x^9/5) - 6/(125 x^14/5)
= 2 - 1/80 - 1/6400 - 3/1024000
= 1.98734082

Real value: 1.987340755