Question

If \[\omega \] is a complex \[n{\rm{th}}\] root of unity, then \[\sum\limits_{r = 1}^n {\left( {ar + b} \right){\omega ^{r - 1}}} \] is equal to
(a) \[\dfrac{{n\left( {n + 1} \right)a}}{2}\]
(b) \[\dfrac{{nb}}{{1 - n}}\]
(c) \[\dfrac{{na}}{{\omega - 1}}\]
(d) None of these

Answer 1

Hint:
Here, we will expand this sum, and factor out \[a\] and \[b\] from the terms. Then, we will simplify the sum further by using the formula for the sum of terms of a geometric progression. Finally, we will use the given information to simplify the expression and obtain the required value of the sum.
Formula Used: We will use the formula for the sum of \[n\] terms of a G.P. is given by the formula \[\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\], where \[a\] is the first term, \[r\] is the common ratio, and \[n\] is the number of terms of the G.P.

Complete step by step solution:
Let the sum \[\sum\limits_{r = 1}^n {\left( {ar + b} \right){\omega ^{r - 1}}} \] be \[S\].
Expanding the sum, we get
\[S = \left( {a \times 1 + b} \right){\omega ^{1 - 1}} + \left( {a \times 2 + b} \right){\omega ^{2 - 1}} + \left( {a \times 3 + b} \right){\omega ^{3 - 1}} + \ldots \ldots \ldots + \left( {a \times n + b} \right){\omega ^{n - 1}}\]
Simplifying the expansion, we get
\[\begin{array}{l} \Rightarrow S = \left( {a + b} \right){\omega ^0} + \left( {2a + b} \right){\omega ^1} + \left( {3a + b} \right){\omega ^2} + \ldots \ldots \ldots + \left( {an + b} \right){\omega ^{n - 1}}\\ \Rightarrow S = \left( {a + b} \right)1 + \left( {2a + b} \right)\omega + \left( {3a + b} \right){\omega ^2} + \ldots \ldots \ldots + \left( {an + b} \right){\omega ^{n - 1}}\\ \Rightarrow S = \left( {a + b} \right) + \left( {2a + b} \right)\omega + \left( {3a + b} \right){\omega ^2} + \ldots \ldots \ldots + \left( {an + b} \right){\omega ^{n - 1}}\end{array}\]
Multiplying the terms using the distributive law of multiplication, we get
\[ \Rightarrow S = \left( {a + b} \right) + \left( {2a\omega + b\omega } \right) + \left( {3a{\omega ^2} + b{\omega ^2}} \right) + \ldots \ldots \ldots + \left( {an{\omega ^{n - 1}} + b{\omega ^{n - 1}}} \right)\]
Rewriting the equation by rearranging the terms, we get
\[ \Rightarrow S = \left( {a + 2a\omega + 3a{\omega ^2} + \ldots \ldots \ldots + an{\omega ^{n - 1}}} \right) + \left( {b + b\omega + b{\omega ^2} \ldots \ldots \ldots + b{\omega ^{n - 1}}} \right)\]
Factoring out \[a\] and \[b\] from the expansion, we can simplify the expansion as
\[ \Rightarrow S = a\left( {1 + 2\omega + 3{\omega ^2} + \ldots \ldots \ldots + n{\omega ^{n - 1}}} \right) + b\left( {1 + \omega + {\omega ^2} \ldots \ldots \ldots + {\omega ^{n - 1}}} \right)\]
Let \[1 + 2\omega + 3{\omega ^2} + \ldots \ldots \ldots + n{\omega ^{n - 1}}\] be \[{S_1}\] and \[1 + \omega + {\omega ^2} \ldots \ldots \ldots + {\omega ^{n - 1}}\] be \[{S_2}\].
Thus, we get
\[ \Rightarrow S = a{S_1} + b{S_2} \ldots \ldots \ldots \left( 1 \right)\]
We will simplify \[{S_1}\] first.
We have
\[ \Rightarrow {S_1} = 1 + 2\omega + 3{\omega ^2} + \ldots \ldots \ldots + n{\omega ^{n - 1}}\]
Multiplying both sides of the equation by \[\omega \], we get
\[\begin{array}{l} \Rightarrow {S_1}\omega = \omega + 2{\omega ^2} + 3{\omega ^3} + \ldots \ldots \ldots + n{\omega ^n}\\ \Rightarrow {S_1}\omega = \omega + 2{\omega ^2} + 3{\omega ^3} + \ldots \ldots \ldots + \left( {n - 1} \right){\omega ^{n - 1}} + n{\omega ^n}\end{array}\]
Subtracting the equation \[{S_1}\omega = \omega + 2{\omega ^2} + 3{\omega ^3} + \ldots \ldots \ldots + \left( {n - 1} \right){\omega ^{n - 1}} + n{\omega ^n}\] from the equation \[{S_1} = 1 + 2\omega + 3{\omega ^2} + \ldots \ldots \ldots + n{\omega ^{n - 1}}\], we get
\[ \Rightarrow {S_1} - {S_1}\omega = 1 + 2\omega + 3{\omega ^2} + \ldots \ldots \ldots + n{\omega ^{n - 1}} - \left( {\omega + 2{\omega ^2} + 3{\omega ^3} + \ldots \ldots \ldots + \left( {n - 1} \right){\omega ^{n - 1}} + n{\omega ^n}} \right)\]
Thus, we get
\[\begin{array}{l} \Rightarrow {S_1}\left( {1 - \omega } \right) = 1 + \omega + {\omega ^2} + \ldots \ldots \ldots + \left[ {n - \left( {n - 1} \right)} \right]{\omega ^{n - 1}} - n{\omega ^n}\\ \Rightarrow {S_1}\left( {1 - \omega } \right) = 1 + \omega + {\omega ^2} + \ldots \ldots \ldots + \left[ {n - n + 1} \right]{\omega ^{n - 1}} - n{\omega ^n}\\ \Rightarrow {S_1}\left( {1 - \omega } \right) = 1 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}} - n{\omega ^n}\end{array}\]
The terms in the sum \[1 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}}\] form a geometric progression, where the first term is 1, the common ratio is \[\omega \], and the number of terms is \[n\].
Substituting \[n = n\], \[r = \omega \], and \[a = 1\] in the formula for sum of terms of a G.P., \[\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\], we get
\[\begin{array}{l} \Rightarrow 1 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}} = \dfrac{{1\left( {{\omega ^n} - 1} \right)}}{{\omega - 1}}\\ \Rightarrow 1 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}} = \dfrac{{{\omega ^n} - 1}}{{\omega - 1}} \ldots \ldots \ldots \left( 2 \right)\end{array}\]
Substituting \[1 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}} = \dfrac{{{\omega ^n} - 1}}{{\omega - 1}}\] in the equation \[{S_1}\left( {1 - \omega } \right) = 1 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}} - n{\omega ^n}\], we get
\[ \Rightarrow {S_1}\left( {1 - \omega } \right) = \dfrac{{{\omega ^n} - 1}}{{\omega - 1}} - n{\omega ^n}\]
It is given that \[\omega \] is a complex \[n{\rm{th}}\] root of unity.
Therefore, we get
\[\omega = \sqrt[n]{1}\]
Thus, \[{\omega ^n} = 1\].
Substituting \[{\omega ^n} = 1\] in the equation, we get
\[\begin{array}{l} \Rightarrow {S_1}\left( {1 - \omega } \right) = \dfrac{{1 - 1}}{{\omega - 1}} - n\left( 1 \right)\\ \Rightarrow {S_1}\left( {1 - \omega } \right) = \dfrac{0}{{\omega - 1}} - n\end{array}\]
Thus, we get
\[\begin{array}{l} \Rightarrow {S_1}\left( {1 - \omega } \right) = 0 - n\\ \Rightarrow {S_1}\left( {1 - \omega } \right) = - n\end{array}\]
Dividing both sides of the equation by \[1 - \omega \], we get
\[\begin{array}{l} \Rightarrow \dfrac{{{S_1}\left( {1 - \omega } \right)}}{{1 - \omega }} = \dfrac{{ - n}}{{1 - \omega }}\\ \Rightarrow {S_1} = \dfrac{{ - n}}{{1 - \omega }}\end{array}\]
Now, we will find the sum \[{S_2}\].
We have
\[{S_2} = 1 + \omega + {\omega ^2} \ldots \ldots \ldots + {\omega ^{n - 1}}\]
Substituting \[1 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}} = \dfrac{{{\omega ^n} - 1}}{{\omega - 1}}\] from equation \[\left( 2 \right)\], we get
\[ \Rightarrow {S_2} = \dfrac{{{\omega ^n} - 1}}{{\omega - 1}}\]
Substituting \[{\omega ^n} = 1\] in the equation, we get
\[\begin{array}{l} \Rightarrow {S_2} = \dfrac{{1 - 1}}{{\omega - 1}}\\ \Rightarrow {S_2} = \dfrac{0}{{\omega - 1}}\end{array}\]
Thus, we get
\[ \Rightarrow {S_2} = 0\]
Now, we can find the sum \[S\].
Substituting \[{S_1} = \dfrac{{ - n}}{{1 - \omega }}\] and \[{S_2} = 0\] in equation \[\left( 1 \right)\], we get
\[ \Rightarrow S = a\left( {\dfrac{{ - n}}{{1 - \omega }}} \right) + b\left( 0 \right)\]
Multiplying the terms of the expression, we get
\[ \Rightarrow S = \dfrac{{ - an}}{{1 - \omega }} + 0\]
Therefore, we get
\[ \Rightarrow S = \dfrac{{ - an}}{{1 - \omega }}\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow S = \dfrac{{ - na}}{{ - \left( {\omega - 1} \right)}}\\ \Rightarrow S = \dfrac{{na}}{{\omega - 1}}\end{array}\]
\[\therefore \] We get the value of the sum \[\sum\limits_{r = 1}^n {\left( {ar + b} \right){\omega ^{r - 1}}} \] as \[\dfrac{{na}}{{\omega - 1}}\].

Thus, the correct option is option (c).

Note:
We have used the distributive law of multiplication in the solution. The distributive law of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\]. Here, the sum is in geometric progression. Geometric progression is a sequence or series in which the two consecutive differences are by a common ratio. However, arithmetic progression is a series or sequence in which there is a common difference between two consecutive numbers.