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Question: If \[\omega \] is a complex \[n{\rm{th}}\] root of unity, then \[\sum\limits_{r = 1}^n {\left( {ar +...

If ω\omega is a complex nthn{\rm{th}} root of unity, then ∑r=1n(ar+b)ωr−1\sum\limits_{r = 1}^n {\left( {ar + b} \right){\omega ^{r - 1}}} is equal to
(a) n(n+1)a2\dfrac{{n\left( {n + 1} \right)a}}{2}
(b) nb1−n\dfrac{{nb}}{{1 - n}}
(c) naω−1\dfrac{{na}}{{\omega - 1}}
(d) None of these

Explanation

Solution

Here, we will expand this sum, and factor out aa and bb 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 nn terms of a G.P. is given by the formula a(rn−1)r−1\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}, where aa is the first term, rr is the common ratio, and nn is the number of terms of the G.P.

Complete step by step solution:
Let the sum ∑r=1n(ar+b)ωr−1\sum\limits_{r = 1}^n {\left( {ar + b} \right){\omega ^{r - 1}}} be SS.
Expanding the sum, we get
S=(a×1+b)ω1−1+(a×2+b)ω2−1+(a×3+b)ω3−1+………+(a×n+b)ωn−1S = \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
⇒S=(a+b)ω0+(2a+b)ω1+(3a+b)ω2+………+(an+b)ωn−1 ⇒S=(a+b)1+(2a+b)ω+(3a+b)ω2+………+(an+b)ωn−1 ⇒S=(a+b)+(2a+b)ω+(3a+b)ω2+………+(an+b)ωn−1\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
⇒S=(a+b)+(2aω+bω)+(3aω2+bω2)+………+(anωn−1+bωn−1)\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
⇒S=(a+2aω+3aω2+………+anωn−1)+(b+bω+bω2………+bωn−1)\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 aa and bb from the expansion, we can simplify the expansion as
⇒S=a(1+2ω+3ω2+………+nωn−1)+b(1+ω+ω2………+ωn−1)\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+2ω+3ω2+………+nωn−11 + 2\omega + 3{\omega ^2} + \ldots \ldots \ldots + n{\omega ^{n - 1}} be S1{S_1} and 1+ω+ω2………+ωn−11 + \omega + {\omega ^2} \ldots \ldots \ldots + {\omega ^{n - 1}} be S2{S_2}.
Thus, we get
⇒S=aS1+bS2………(1)\Rightarrow S = a{S_1} + b{S_2} \ldots \ldots \ldots \left( 1 \right)
We will simplify S1{S_1} first.
We have
⇒S1=1+2ω+3ω2+………+nωn−1\Rightarrow {S_1} = 1 + 2\omega + 3{\omega ^2} + \ldots \ldots \ldots + n{\omega ^{n - 1}}
Multiplying both sides of the equation by ω\omega , we get
⇒S1ω=ω+2ω2+3ω3+………+nωn ⇒S1ω=ω+2ω2+3ω3+………+(n−1)ωn−1+nωn\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 S1ω=ω+2ω2+3ω3+………+(n−1)ωn−1+nωn{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 S1=1+2ω+3ω2+………+nωn−1{S_1} = 1 + 2\omega + 3{\omega ^2} + \ldots \ldots \ldots + n{\omega ^{n - 1}}, we get
⇒S1−S1ω=1+2ω+3ω2+………+nωn−1−(ω+2ω2+3ω3+………+(n−1)ωn−1+nωn)\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
⇒S1(1−ω)=1+ω+ω2+………+[n−(n−1)]ωn−1−nωn ⇒S1(1−ω)=1+ω+ω2+………+[n−n+1]ωn−1−nωn ⇒S1(1−ω)=1+ω+ω2+………+ωn−1−nωn\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+ω+ω2+………+ωn−11 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}} form a geometric progression, where the first term is 1, the common ratio is ω\omega , and the number of terms is nn.
Substituting n=nn = n, r=ωr = \omega , and a=1a = 1 in the formula for sum of terms of a G.P., a(rn−1)r−1\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}, we get
⇒1+ω+ω2+………+ωn−1=1(ωn−1)ω−1 ⇒1+ω+ω2+………+ωn−1=ωn−1ω−1………(2)\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+ω+ω2+………+ωn−1=ωn−1ω−11 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}} = \dfrac{{{\omega ^n} - 1}}{{\omega - 1}} in the equation S1(1−ω)=1+ω+ω2+………+ωn−1−nωn{S_1}\left( {1 - \omega } \right) = 1 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}} - n{\omega ^n}, we get
⇒S1(1−ω)=ωn−1ω−1−nωn\Rightarrow {S_1}\left( {1 - \omega } \right) = \dfrac{{{\omega ^n} - 1}}{{\omega - 1}} - n{\omega ^n}
It is given that ω\omega is a complex nthn{\rm{th}} root of unity.
Therefore, we get
ω=1n\omega = \sqrt[n]{1}
Thus, ωn=1{\omega ^n} = 1.
Substituting ωn=1{\omega ^n} = 1 in the equation, we get
⇒S1(1−ω)=1−1ω−1−n(1) ⇒S1(1−ω)=0ω−1−n\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
⇒S1(1−ω)=0−n ⇒S1(1−ω)=−n\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−ω1 - \omega , we get
⇒S1(1−ω)1−ω=−n1−ω ⇒S1=−n1−ω\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 S2{S_2}.
We have
S2=1+ω+ω2………+ωn−1{S_2} = 1 + \omega + {\omega ^2} \ldots \ldots \ldots + {\omega ^{n - 1}}
Substituting 1+ω+ω2+………+ωn−1=ωn−1ω−11 + \omega + {\omega ^2} + \ldots \ldots \ldots + {\omega ^{n - 1}} = \dfrac{{{\omega ^n} - 1}}{{\omega - 1}} from equation (2)\left( 2 \right), we get
⇒S2=ωn−1ω−1\Rightarrow {S_2} = \dfrac{{{\omega ^n} - 1}}{{\omega - 1}}
Substituting ωn=1{\omega ^n} = 1 in the equation, we get
⇒S2=1−1ω−1 ⇒S2=0ω−1\begin{array}{l} \Rightarrow {S_2} = \dfrac{{1 - 1}}{{\omega - 1}}\\\ \Rightarrow {S_2} = \dfrac{0}{{\omega - 1}}\end{array}
Thus, we get
⇒S2=0\Rightarrow {S_2} = 0
Now, we can find the sum SS.
Substituting S1=−n1−ω{S_1} = \dfrac{{ - n}}{{1 - \omega }} and S2=0{S_2} = 0 in equation (1)\left( 1 \right), we get
⇒S=a(−n1−ω)+b(0)\Rightarrow S = a\left( {\dfrac{{ - n}}{{1 - \omega }}} \right) + b\left( 0 \right)
Multiplying the terms of the expression, we get
⇒S=−an1−ω+0\Rightarrow S = \dfrac{{ - an}}{{1 - \omega }} + 0
Therefore, we get
⇒S=−an1−ω\Rightarrow S = \dfrac{{ - an}}{{1 - \omega }}
Simplifying the expression, we get
⇒S=−na−(ω−1) ⇒S=naω−1\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 ∑r=1n(ar+b)ωr−1\sum\limits_{r = 1}^n {\left( {ar + b} \right){\omega ^{r - 1}}} as naω−1\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(b+c)=a⋅b+a⋅ca\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.