Question

Need help with BINOMIALS

Answer 1

Simplify this.
\[
\left(2x-\frac{1}{2x}\right)^6=\sum_{k=0}^6\binom{6}{k}(2x)^k\frac{1}{(2x)^{6-k}}
\]

Answer 2

I don't understand yours but I guess it can also be written as \[\sum_{r=0}^{6} (2x)^{6-r} (\frac{ 1 }{ 2x })^r\] then?

Answer 3

And r=0 as you have a \(x^2\) in the first term.

Answer 4

Not r=0 but r=3.

Answer 5

still can't get it.. Could you show me the full working please?

Answer 6

I don't mind if you write. Typing is too tedious

Answer 7

\[
\text{We are interested in these two terms of the binomial series}:\\
r=2 \text{ as }1\times\binom{6}{2}(2x)^{6-2}\left(\frac{1}{2x}\right)^2=60x^2\\
r=3\text{ as } x^2\times\binom{6}{3}(2x)^{6-3}\left(\frac{1}{2x}\right)^{3}=20x^2
\]
The \(1\) and \(x^2\) comes from the first bracket.

Answer 8

Oh pellet... I left out the -1. Let me do this again.

Answer 9

The binomial series for the second bracket is:
\[
\sum_{r=0}^6(2x)^{6-r}(-1)^r\frac{1}{(2x)^r}=\sum_{r=0}^6(2x)^{6-2r}(-1)^r
\]
We get rid of the first bracket.
\[
\begin{align*}
(1+x^2)\left(2x-\frac{1}{2x}\right)^6&=(1+x^2)\left(\sum_{r=0}^6(2x)^{6-2r}(-1)^r
\right)\\
&=\sum_{r=0}^6(2x)^{6-2r}(-1)^r
+x^2\sum_{r=0}^6(2x)^{6-2r}(-1)^r
\end{align*}
\]
Now try solve this. We only want \(x^2\) terms.

Answer 10

Got it! 20. Thanks

Answer 11

40*