Question

If \[{S_n} = \sum\nolimits_n^{r = 0} {\dfrac{1}{{{}^n{C_r}}}} \] and \[{t_n} = \sum\nolimits_n^{r = 0} {\dfrac{r}{{{}^n{C_r}}}} \], then \[\dfrac{{{t_n}}}{{{S_n}}}\] is equal to
A) \[\dfrac{1}{2}n\]
B) \[\dfrac{1}{2}n - 1\]
C) \[n - 1\]
D) \[\dfrac{{2n - 1}}{2}\]

Answer 1

Hint:
Here, we will first consider only the relation for arithmetic progression. We will then introduce a variable in the numerator and simplify it further. We will then substitute the given equation of arithmetic series in the equation and simplify it further to the ratio of the arithmetic progression to the arithmetic series.

Complete step by step solution:
We are given that \[{S_n} = \sum\nolimits_n^{r = 0} {\dfrac{1}{{{}^n{C_r}}}} \] and \[{t_n} = \sum\nolimits_n^{r = 0} {\dfrac{r}{{{}^n{C_r}}}} \]
Now, we will first consider only the relation for arithmetic progression \[{t_n} = \sum\nolimits_n^{r = 0} {\dfrac{r}{{{}^n{C_r}}}} \].
Now by adding and subtracting the term \[n\], in the numerator we get
\[ \Rightarrow {t_n} = \sum\nolimits_n^{r = 0} {\dfrac{{n - n + r}}{{{}^n{C_r}}}} \]
By rewriting the equation, we get
\[ \Rightarrow {t_n} = \sum\nolimits_n^{r = 0} {\dfrac{{n - \left( {n - r} \right)}}{{{}^n{C_r}}}} \]
By separating the variables in the numerator, we get
\[ \Rightarrow {t_n} = \sum\nolimits_n^{r = 0} {\left[ {\dfrac{n}{{{}^n{C_r}}} - \dfrac{{\left( {n - r} \right)}}{{{}^n{C_r}}}} \right]} \]
By multiplying the summation term to the terms inside the brackets, we get\[ \Rightarrow {t_n} = \sum\nolimits_n^{r = 0} {\dfrac{n}{{{}^n{C_r}}} - \sum\nolimits_n^{r = 0} {\dfrac{{\left( {n - r} \right)}}{{{}^n{C_r}}}} } \]
By replacing the variable \[n - r\] as \[r\] in the above equation, we get
\[ \Rightarrow {t_n} = n\sum\nolimits_n^{r = 0} {\dfrac{1}{{{}^n{C_r}}} - \sum\nolimits_n^{r = 0} {\dfrac{r}{{{}^n{C_r}}}} } \]
By rewriting the equation in terms of the equation of the given arithmetic series and given arithmetic progression, we get
\[ \Rightarrow {t_n} = n{S_n} - {t_n}\]
By rewriting the equation, we get
\[ \Rightarrow {t_n} + {t_n} = n{S_n}\]
\[ \Rightarrow 2{t_n} = n{S_n}\]
Again by rewriting the equation in terms of ratio of the arithmetic progression to the arithmetic ratio, we get
\[ \Rightarrow \dfrac{{{t_n}}}{{{S_n}}} = \dfrac{n}{2}\]
\[ \Rightarrow \dfrac{{{t_n}}}{{{S_n}}} = \dfrac{1}{2}n\]
Therefore, the ratio of \[\dfrac{{{t_n}}}{{{S_n}}}\] is \[\dfrac{1}{2}n\].
Thus Option (A) is the correct answer.

Note:
We know that Arithmetic progression is a sequence of numbers where the difference between the two consecutive numbers is a constant. The arithmetic series is the sum of the terms of the arithmetic progression. An arithmetic sequence of numbers is also defined as a sequence of numbers where each number is the sum of the preceding number and common difference (d) is a constant. If the same number is added or subtracted from each term of an A.P., then the resulting terms in the sequence are also in A.P. but with the same common difference. We should remember that whenever we are adding a variable, then the same variable has to be subtracted as it does not affect the term.