Question 1189036
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I very much think the other tutor completely missed the point of your question....<br>
Yes, you can "multiply out" the expression ((x+5)^(0.5)) using binomial expansion.<br>
Many students will be familiar with binomial expansion when the power is an integer; few students will be familiar with it when the power is not an integer.<br>
Because the exponent is not an integer, the expansion does not terminate; it has an infinite number of terms.  The first few terms will give good approximations.<br>
The key to understanding binomial expansion with non-integer powers is to look at how binomial coefficients are calculated.<br>
For all n, including non-integer values of n...<br>
{{{C(n,0) = 1}}}<br>
For all other binomial coefficients C(n,r), the coefficient can be calculated as a fraction with "r" numbers in both numerator and denominator, starting with n and decrementing by 1 in the numerator, and starting with r and decrementing by 1 in the denominator.  That gives us...<br>
{{{C(n,1) = n/1 = n}}}
{{{C(n,2) = ((n)(n-1))/(2*1)}}}
{{{C(n,3) = ((n)(n-1)(n-2))/(3*2*1)}}}
etc....<br>
For example for n=3, we get what (I hope) are the familiar coefficients for n=3:<br>
{{{C(3,0) = 1}}}
{{{C(3,1) = (3/1) = 3}}}
{{{C(3,2) = ((3*2)/(2*1)) = 6/2 = 3}}}
{{{C(3,3) = ((3*2*1)/(3*2*1)) = 6/6 = 1}}}<br>
And using the same method for calculating the coefficients for n=0.5, we get the unfamiliar coefficients for a binomial expansion to the 0.5 power:<br>
{{{C(0.5,0) = 1}}}
{{{C(0.5,1) = (0.5/1) = 1/2}}}
{{{C(0.5,2) = (((0.5)(-0.5))/(2*1)) = -1/8}}}
{{{C(0.5,3) = (((0.5)(-0.5)(-1.5))/(3*2*1)) = 1/16}}}<br>
Now we can write out the first few terms of the binomial expansion of (x+5)^(-0.5).  We have the binomial coefficients; and from one term to the next the power of x decreases by 1 and the power of 5 increases by 1:<br>
{{{f(x) = (x+5)^(-0.5) = (1)(x^(0.5))+(1/2)(5)(x^(-0.5))-(1/8)(25)(x^(-1.5))+(1/16)(125)(x^(-2.5))}}}<br>
Evaluating that expression for values of x that should produce whole number results, we find<br><pre>

   x  x+5  (x+5)^(0.5)  f(x)  (4 decimal places)
  ------------------------------
   4   9        3       3.1035
  11  16        4       4.0042
  20  25        5       5.0006
  31  36        6       6.0001
  44  49        7       7.0000
  59  64        8       8.0000</pre>
For small values of x, where this approximation is not very accurate, using 2 or 3 more terms of the binomial expansion would improve the accuracy.<br>