Question

How do you use the sigma function to write the sum for $ \dfrac{1}{4} + \dfrac{3}{8} + \dfrac{7}{{16}} + \dfrac{{15}}{{32}} + \dfrac{{31}}{{64}} $ ?

Answer 1

Hint: In order to determine the sigma notation for the sum of given finite series, derive the general term for numerator values of the terms as one less than the successive powers of $ 2 $ and also the general term for the denominator values for the terms as successive powers of $ 2 $ except first term. Now combine both general terms to obtain the general term for the whole series. Now use the sigma from $ n = 1 $ to 5 to obtain the required result.

Complete step by step solution:
We are given a finite sum series as $ \dfrac{1}{4} + \dfrac{3}{8} + \dfrac{7}{{16}} + \dfrac{{15}}{{32}} + \dfrac{{31}}{{64}} $ . And we have to derive the sigma or summation function for this.
Let first find out what is summation or sigma function and its representation.
Summation or Sigma function is basically a representation of the given series in a short and compact form. To represent the sum, a Greek capital letter $ \sum {} $ is used which is generally called summation.
 Standard summation notation is $ \sum\limits_{i = 1}^n {{x_i}} $ where $ i $ is the index of summation, \[n\] is the stopping point or upper limit of summation and $ {x_i} $ is the typical element.
If we look at the denominator of the terms in the given sum series, we have
 $ \left( {4,8,16,32,64} \right) $ ,
We can clearly see that these numbers are successive powers of $ 2 $ in which first term $ 2 $ is missing. In general form, we can write
 $ {2^{n + 1}} $ where $ n \in \left\{ {1,2,3,4,5} \right\} $ ---(1)
Now similarly if we look at the sequence for the values of numerators in the terms
 $ \left( {1,3,7,15,31} \right) $ , we can see that these terms are one less than successive powers of 2. So in general term, they can be written as
 $ {2^n} - 1 $ where $ n \in \left\{ {1,2,3,4,5} \right\} $ ---------(2)
If we combine the equations (1) and (2) , the general term for the series will be
 $ {u_n} = \dfrac{{{2^n} - 1}}{{{2^{n + 1}}}} $ where $ n \in \left\{ {1,2,3,4,5} \right\} $
Hence , we can write the finite sum series using sigma notation as
 $ \dfrac{1}{4} + \dfrac{3}{8} + \dfrac{7}{{16}} + \dfrac{{15}}{{32}} + \dfrac{{31}}{{64}} \Rightarrow \sum\limits_{n = 1}^5 {\dfrac{{{2^n} - 1}}{{{2^{n + 1}}}}} $
Therefore, the sigma notation of the given sum series is $ \sum\limits_{n = 1}^5 {\dfrac{{{2^n} - 1}}{{{2^{n + 1}}}}} $ .
So, the correct answer is “ $ \sum\limits_{n = 1}^5 {\dfrac{{{2^n} - 1}}{{{2^{n + 1}}}}} $ ”.

Note: The sigma notation is a very compact and easy way of writing the sum of a series of terms.
Below the sigma symbol, we write the starting value of the variable in general term and above we mention the ending value of general term.
Be careful while writing the general term of the series in order to correctly represent the terms of the series of sum .