# [SOLVED] Binomial expansion problem

• February 22nd 2010, 09:41 AM
dojo
[SOLVED] Binomial expansion problem
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

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??
• February 22nd 2010, 11:05 AM
Hello dojo
Quote:

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

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 $x^2$, whence it becomes $\big(x^2\big)^{18}$:
$\left(x^2-\frac2x\right)^{18} = x^{36}\left(1-\frac{2}{x^3}\right)^{18}$
Now let $y = -\frac{2}{x^3}$, and expand $(1+y)^{18}$, the general term of which is $\binom{18}{r}y^r$, to get the general term:
$x^{36}\binom{18}{r}\left(-\frac{2}{x^3}\right)^r=(-2)^r\binom{18}{r}x^{36-3r}$
The term that's independent of $x$ is the one where the power of $x$ is zero. So put $r = 12$ ...

(ii)
$(1+x)^{16} = 1 + 16x+ \binom{16}{2}x^2 + \binom{16}{3}x^3+ ...$
Now put $x = z+z^2$:
$(1+z+z^2)^{16} = 1 + 16(z+z^2)+ \binom{16}{2}(z+z^2)^2 + \binom{16}{3}(z+z^2)^3+ ...$
= ...
Expand each $(z+z^2)$ term up to the term in $z^3$; collect like terms, and you're done.

• February 22nd 2010, 12:18 PM
Soroban
Hello, dojo!

Quote:

(i) Find the general term in the expansion of: . $\left(x^2 - \frac{2}{x}\right)^{18}$
Assuming that you know the Binomial Theorem, this is easy:

. . ${18\choose n}\left(x^2\right)^n\left(\text{-}\frac{2}{x}\right)^{18-n}\;\;\text{ for }n \,=\,18,17,16\,\hdots 0$

Quote:

Hence, find the term which is independant of $x$.
${\color{blue}\text{It means "Find the term which has no }x."}$

Somewhere in the middle, we have: . ${18\choose6}\left(x^2\right)^6\left(\text{-}\frac{2}{x}\right)^{12}$

. . $=\; \frac{18!}{6!\,12!}\left(x^{12}\right)\left(\frac{ (\text{-}2)^{12}}{x^{12}}\right) \;=\;(18,\!564)\left(x^{12}\right)\left(\frac{4,\! 096}{x^{12}}\right) \;=\; 76,\!038,\!144$

• February 22nd 2010, 12:41 PM
dojo
Thank you both for the insight very much appreciated!