Binomial expression help!

**Confuzzled?** · January 30th 2006, 04:25 AM

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

**CaptainBlack** · January 30th 2006, 05:55 AM

Originally Posted by Confuzzled?

Find the first three terms in these expansions in ascending powers of x
a) (x+1/x)^6

$(x+1/x)^6=(1/x)^6+6(1/x)^5x+\frac{6\times 5}{2}(1/x)^4 x^2+...$

$(x+1/x)^6=x^{-6}+6x^{-4}+15x^{-2}+...$

RonL

**CaptainBlack** · January 30th 2006, 06:08 AM

Originally Posted by Confuzzled?

b) In the expansion of (1+x/2)^n in ascending powers of x the coefficent of x^2 is 30. Find n.

$(1+x/2)^n =1+n(x/2)+\frac{n(n-1)}{2}(x/2)^2+...$

So the coeff of $x^2$ in the expansion is:

$\frac{n(n-1)}{8}=30$ ,

which has solutions $n=-15$ or $n=16$ .

I presume there is an implicit assumption that $n \ge 1$ so
the answer would be $n=16$ .

RonL

**Confuzzled?** · January 30th 2006, 07:04 AM

Thank you so much!

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