Question

If nth root of unity of a complex number is given as $\omega$, 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) $\dfrac{ab}{1-n}$
(c) $\dfrac{na}{\omega -1}$
(d) None of these.

Answer 1

Here, first we have to expand the given sequence and find what type of progression (Arithmetic Progression or Geometric Progression) we have, simplify the series using basic mathematical operations. Use the formula, of geometric progression ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ and the given statement saying ${{\omega }^{n}}=1$ to further simplify the series. In the second part, multiply the series by $\omega$ and subtract it from the obtained expression and simplify it further to find the required result.

Complete step-by-step solution:
Here, we have a series of summation of $\left( ar+b \right){{\omega }^{r-1}}$ from $r=1$ to $n$, let us first expand the given sequence and find the series. We need to substitute the value of r from 1 to $n$ with a ‘+’ sign after every term. Also, we have been given, $\omega$ is a complex nth root of unity which means ${{\omega }^{n}}=1$.
${{S}_{n}}$= $\left[ a\left( 1 \right)+b \right]{{\omega }^{1-1}}+\left[ a\left( 2 \right)+b \right]{{\omega }^{2-1}}+\left[ a\left( 3 \right)+b \right]{{\omega }^{3-1}}+........+\left[ a\left( n \right)+b \right]{{\omega }^{n-1}}$
Let us simplify the above series by using basic mathematical operations and basic principles of indices, we get
${{S}_{n}}$= $\left( a+b \right)+\left( 2a+b \right)\omega +\left( 3a+b \right){{\omega }^{2}}+\left( 4a+b \right){{\omega }^{3}}+...........+\left( na+b \right){{\omega }^{n-1}}$
In the above expression, let us group all the $a$ terms and $b$ terms separately, we get
${{S}_{n}}=\left( a+2a\omega +3a{{\omega }^{2}}+4a{{\omega }^{3}}+....+na{{\omega }^{n-1}} \right)+\left( b+b\omega +b{{\omega }^{2}}+b{{\omega }^{3}}+.....+b{{\omega }^{n-1}} \right)$
Now, let us take $a$ and $b$as common from the above expression and simplify, we get
${{S}_{n}}=a\left( 1+2\omega +3{{\omega }^{2}}+......+n{{\omega }^{n-1}} \right)+b\left( 1+\omega +{{\omega }^{2}}+{{\omega }^{3}}+......+{{\omega }^{n-1}} \right)$………… (1)
Now, we can see that $\left( 1+\omega +{{\omega }^{2}}+{{\omega }^{3}}+......+{{\omega }^{n-1}} \right)$ is in the form of geometric progression and we know that, $S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ for geometric progression where we have first term, $a=1$ and common ratio, $r=\omega$, we get
$S=\dfrac{1\left( {{\omega }^{n}}-1 \right)}{\omega -1}$
Since, we know ${{\omega }^{n}}=1$, we get
$\begin{aligned} & S=\dfrac{1\left( 1-1 \right)}{\omega -1} \\\ & =0 \end{aligned}$
Therefore, from equation (1) and the above equation, we get
${{S}_{n}}=a\left( 1+2\omega +3{{\omega }^{2}}+......+n{{\omega }^{n-1}} \right)+b\left( 0 \right)$
$\Rightarrow{{S}_{n}}=a\left( 1+2\omega +3{{\omega }^{2}}+......+n{{\omega }^{n-1}} \right)$............... (2)
Now, we will multiply equation (2) by $\omega$, we get
$\begin{aligned} & {{S}_{n}}\cdot \omega =a\omega \left( 1+2\omega +3{{\omega }^{2}}+.....+n{{\omega }^{n-1}} \right) \\\ & =a\omega +2a{{\omega }^{2}}+3a{{\omega }^{3}}+.....+na{{\omega }^{n}} \end{aligned}$
Since, ${{\omega }^{1}}\times {{\omega }^{n-1}}={{\omega }^{n-1+1}}={{\omega }^{n}}$.
${{S}_{n}}\cdot \omega =a\omega +2a{{\omega }^{2}}+3a{{\omega }^{3}}+.....+na{{\omega }^{n}}$……………. (3)
Now, let us subtract equation (3) from equation (2), we get
(2) – (3):
${{S}_{n}}-{{S}_{n}}\cdot \omega =a\left( 1+2\omega +3{{\omega }^{2}}+......+n{{\omega }^{n-1}} \right)-a\omega -2a{{\omega }^{2}}-3a{{\omega }^{3}}-.....-na{{\omega }^{n}}$
Now, let us take $a$ variable common from the above expression, we get
${{S}_{n}}\left( 1-\omega \right)=a\left( 1+2\omega +3{{\omega }^{2}}+.....+n{{\omega }^{n-1}}-\omega -2{{\omega }^{2}}-3{{\omega }^{3}}-.....-n{{\omega }^{n}} \right)$
Let us solve the above expression and simplify it further, we get
$\Rightarrow {{S}_{n}}\left( 1-\omega \right)=a\left( 1+2\omega -\omega +3{{\omega }^{2}}-2{{\omega }^{2}}+4{{\omega }^{3}}-3{{\omega }^{3}}+.....+n{{\omega }^{n-1}}-n{{\omega }^{n}} \right)$
$\Rightarrow {{S}_{n}}\left( 1-\omega \right)=a\left( 1+\omega +{{\omega }^{2}}+{{\omega }^{3}}+.....+n{{\omega }^{n-1}} \right)-na{{\omega }^{n}}$
In the above expression, we can see there is a geometric progression, if we use the formula of geometric progression, we will get the value as 0. Therefore, $\left( 1+\omega +{{\omega }^{2}}+{{\omega }^{3}}+......+{{\omega }^{n-1}} \right)$ = 0, when ${{\omega }^{n}}=1$. We get,
${{S}_{n}}\left( 1-\omega \right)=a\left( 0 \right)-na{{\omega }^{n}}$
$\Rightarrow {{S}_{n}}\left( 1-\omega \right)=0-na\left( 1 \right)$
$\Rightarrow {{S}_{n}}=\dfrac{-na}{\left( 1-\omega \right)}$
Multiplying by – 1 on the numerator and denominator on the right-hand side of the equation, we get
${{S}_{n}}=\dfrac{na}{\left( \omega -1 \right)}$
Therefore, the value of the sequence is $\dfrac{na}{\left( \omega -1 \right)}$.

Note: In this question, we need to understand the basics of indices, the one formula of indices which is used in this particular question is ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. Geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.