Binomial expansion

Hello everyone,

I'm stuck on this question - when I substitute 0.001 into the expanded form, I get 0.997991.... , which is ten times smaller than the actual answer.

I'm just wondering where I went wrong - did I do the binomial expansion wrong?

Hello everyone,

I'm stuck on this question - when I substitute 0.001 into the expanded form, I get 0.997991.... , which is ten times smaller than the actual answer.

I'm just wondering where I went wrong - did I do the binomial expansion wrong?

Firstly note this:
Sub $x = 0.001$, this gives us $\sqrt[5]{1-0.01} = \sqrt[5]{0.99}$.

Then,
check $\sqrt[5]{99000} = \sqrt[5]{0.99 \times 100000} = \sqrt[5]{0.99} \times \sqrt[5]{100000}$

This gives us: $10 \sqrt[5]{0.99}$

So the answer you are after is indeed 10 times $\sqrt[5]{0.99}$

In part (b) it looks as though you have copied the first bracketed expression wrongly, with 3 rather than 27 in the denominator of two of the terms. When it comes to multiplying out the two brackets, the approach I would recommend is:

* On one row, multiply the first expression through by the first term of the second expression.

* Start a new row. Multiply the first expression through by the second term of the second expression, lining up column-wise with the first row so the x^1 terms form a column and the x^2 terms form a column. You can stop at the x^2 term.

* Start a third row and do likewise for the third term in the second expression.

* Now you can add up the columns to find the coefficients of x^0, x^1 and x^2.

I find this approach cuts down errors (and it will help the examiner to see what you're doing).

So how do you find where x is valid for this expression?

You have two intervals of validity for x. The smallest one wins.

so would I be right in saying |x| < 1.5?

The overall validity is set by whichever is the more restrictive of the individual validities.