M17 Q05

The value of \(\frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20}\) is between?

A. \(\frac{1}{2}\) and \(\frac{2}{3}\)
B. \(\frac{2}{3}\) and \(\frac{3}{4}\)
C. \(\frac{3}{4}\) and \(\frac{9}{10}\)
D. \(\frac{9}{10}\) and \(\frac{10}{9}\)
E. \(\frac{10}{9}\) and \(\frac{3}{2}\)

We have the sum of a geometric progression with the first term equal to \(\frac{1}{2}\) and the common ratio also equal to \(\frac{1}{2}\).

Now, the sum of infinite geometric progression with common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term. So, if we had infinite geometric progression instead of just 20 terms then its sum would be \(Sum=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1\). Which means that the sum of this sequence will never exceed 1, also as we have big enough number of terms (20) then the sum will be very close to 1, so we can safely choose answer choice D.

Answer: D.

One can also use direct formula.
We have geometric progression with \(b=\frac{1}{2}\), \(r=\frac{1}{2}\) and \(n=20\);

\(S_n=\frac{b(1-r^n)}{(1-r)}\) --> \(S_{20}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{(1-\frac{1}{2})}=1-\frac{1}{2^{20}}\). Since \(\frac{1}{2^{20}}\) is very small number then \(1-\frac{1}{2^{20}}\) will be less than 1 but very close to it.

Answer: D.

Thanks everyone. The methods are just amazing.

Bunuel, your direct formula method is fantastic.

Thanks & Regards
Vinni