Question

If ${S_n} = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + ... + \dfrac{1}{{{2^{n - 1}}}}$ , find the least value of $n$ for which $2 - {S_n} < \dfrac{1}{{100}}$ ?

Answer 1

Hint: Here we are asked to find the least value of the term $n$ so that $2 - {S_n} < \dfrac{1}{{100}}$ , where ${S_n}$ given. For that, we first need to find whether the given sum of series is AP or GP. If it is GP then we can use the general formula of GP to find the sum of the geometric progression. Then we will substitute it in the given condition to find the least value of the term $n$.
Formula: Formula we need to know:
The sum of GP $a + ar + a{r^2} + a{r^3} + ... + a{r^n}$ : ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$
$a - $ the first term in the GP
$r - $ the common difference in the GP

Complete step by step answer:
It is given that ${S_n} = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + ... + \dfrac{1}{{{2^{n - 1}}}}$ , we aim to find the least value of the term $n$ for which $2 - {S_n} < \dfrac{1}{{100}}$ .
Let us consider the given series, ${S_n} = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + ... + \dfrac{1}{{{2^{n - 1}}}}$ as we can see that this is a geometric progression that has the common ratio $r = \dfrac{1}{2}$ and the first term $a = 1$ .
Thus, using the formula of the sum of the GP we get the sum of the given series as
${S_n} = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + ... + \dfrac{1}{{{2^{n - 1}}}} = \dfrac{{1\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^n}} \right)}}{{1 - \dfrac{1}{2}}}$
Now on simplifying the above we get
$ \Rightarrow {S_n} = \dfrac{{1 - {{\left( {\dfrac{1}{2}} \right)}^n}}}{{1 - \dfrac{1}{2}}}$
On further simplification we get
$ \Rightarrow {S_n} = \dfrac{{1 - \dfrac{1}{{{2^n}}}}}{{\dfrac{1}{2}}}$
$ \Rightarrow {S_n} = 2\left( {1 - \dfrac{1}{{{2^n}}}} \right)$
$ \Rightarrow {S_n} = 2 - {2.2^{ - n}}$
$ \Rightarrow {S_n} = 2 - {2^{1 - n}}$
Thus, we have got the value of the sum of the series. Now let us substitute it in the given condition to find the least value of the term $n$ .
Consider the given condition $2 - {S_n} < \dfrac{1}{{100}}$ now substituting the value of ${S_n}$ we get
$2 - \left( {2 - {2^{1 - n}}} \right) < \dfrac{1}{{100}}$
Let us simplify the above inequality we get
$ \Rightarrow {2^{1 - n}} < \dfrac{1}{{100}}$
We can see that for the values of $n$ greater than seven the condition is satisfied. Thus, the least integer that satisfies the given condition $2 - {S_n} < \dfrac{1}{{100}}$ is $8$ .
That is, $n = 8$ is the least integer for which $2 - {S_n} < \dfrac{1}{{100}}$ , where ${S_n} = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + ... + \dfrac{1}{{{2^{n - 1}}}}$ .

Note:
 The geometric progression is a sequence where the terms are generated in such a way that every succeeding term is obtained by multiplying its preceding term by a constant value. The term $r$ in the formula of the sum of the series denotes the common ratio of the series that is the constant value for any two consecutive terms in the series. The sum of infinite geometric progression can be found by using the formula $\dfrac{a}{{1 - r}}$ .