Hello dojo Originally Posted by

**dojo** I'm having problem with my revision sheets - Any help would be much appreciated...

(i) Write down the general term in the power series expansion of

((x^2) - (2/x))^18

WHAT IS THE GENERAL TERM AND HOW WOULD I GO ABOUT THIS?

Hence find the term which is independant of x

I DONT KNOW WHAT THIS MEANS

(ii)Find the first four terms in the expansion of (1+x)^16 in ascending powers of x.

hence expand (1 + z + z^2) in ascending powers of z up to and including the term in z^3

DO YOU FACTORISE THIS AND RAISE BOTH PARENTHESIS TO THE POWER 16??

(i) Take out a factor $\displaystyle x^2$, whence it becomes $\displaystyle \big(x^2\big)^{18}$:$\displaystyle \left(x^2-\frac2x\right)^{18} = x^{36}\left(1-\frac{2}{x^3}\right)^{18}$

Now let $\displaystyle y = -\frac{2}{x^3}$, and expand $\displaystyle (1+y)^{18}$, the general term of which is $\displaystyle \binom{18}{r}y^r$, to get the general term:$\displaystyle x^{36}\binom{18}{r}\left(-\frac{2}{x^3}\right)^r=(-2)^r\binom{18}{r}x^{36-3r}$

The term that's independent of $\displaystyle x$ is the one where the power of $\displaystyle x$ is zero. So put $\displaystyle r = 12$ ...

(ii) $\displaystyle (1+x)^{16} = 1 + 16x+ \binom{16}{2}x^2 + \binom{16}{3}x^3+ ...$

Now put $\displaystyle x = z+z^2$:$\displaystyle (1+z+z^2)^{16} = 1 + 16(z+z^2)+ \binom{16}{2}(z+z^2)^2 + \binom{16}{3}(z+z^2)^3+ ...$= ...

Expand each $\displaystyle (z+z^2)$ term up to the term in $\displaystyle z^3$; collect like terms, and you're done.

Grandad