Find the first three terms in these expansions in ascending powers of x
a) (x+1/x)^6
b) In the expansion of (1+x/2)^n in ascending powers of x the coefficent of x^2 is 30. Find n.
Thanks in advance if you can help
$\displaystyle (1+x/2)^n =1+n(x/2)+\frac{n(n-1)}{2}(x/2)^2+...$Originally Posted by Confuzzled?
So the coeff of $\displaystyle x^2$ in the expansion is:
$\displaystyle \frac{n(n-1)}{8}=30$,
which has solutions $\displaystyle n=-15$ or $\displaystyle n=16$.
I presume there is an implicit assumption that $\displaystyle n \ge 1$ so
the answer would be $\displaystyle n=16$.
RonL