Alan

As follows:

You can now complete the problem.

(The £ sign in the 6'th line should be a 3 of course!!)

Note that \(log_2(2x)=log_2(2)+log_2(x)\) and \(log_2(2)=1\) 

Alternatively, you can do it the long way by noting that the n'th term of the geometric sequence is given by \(2^{n+2}\).

See https://web2.0calc.com/questions/product_3#r2

Here's one way:

Note that \(2^a2^b2^c....=2^{a+b+c...}\)

In your case \(a+b+c\) forms an infinite geometric series, ie. \(b = ar, c = ar^2, ...etc\)

The limit of the infinite geometric series here is \(\frac{a}{1-r}\) or \(\frac{1/3}{1-1/3}\) 

You can complete the calculation.

As follows, I think:

I think you mean that 18 is the number of degrees of freedom!  Look in some chi-square distribution tables (e.g. https://en.wikibooks.org/wiki/Engineering_Tables/Chi-Squared_Distibution) along the row for DF = 18 until you get to a number close to 21.6, then look at the column heading for the probability.  

(NB: In the reference above the probabilities represent greater than rather than less than, so you would need to subtract the resulting probability from 1 for your case.)

Try this:

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As follows: