Question

1/w +1/w²+1/w³ +...1/w¹⁰⁰​

Answer 1

Given to find,

$\longrightarrow S=\dfrac{1}{\omega}+\dfrac{1}{\omega^2}+\dfrac{1}{\omega^3}+\,\dots\,+\dfrac{1}{\omega^{100}}\quad\qaud\dots(1)$

where $\omega$ is cube root of unity but $\omega\neq1,$ so that,

$\longrightarrow\omega^3=1$

$\longrightarrow\omega^3-1=0$

Since $a^3-1=(a-1)(a^2+a+1),$

$\longrightarrow(\omega-1)(\omega^2+\omega+1)=0$

But $\omega\neq1.$ So we get,

$\longrightarrow\omega^2+\omega+1=0$

Taking $\omega^3=1$ in (1),

$\longrightarrow S=\dfrac{\omega^3}{\omega}+\dfrac{\omega^3}{\omega^2}+\dfrac{\omega^3}{\omega^3}+\,\dots\,+\dfrac{\omega^3}{\omega^{100}}$

$\longrightarrow S=\omega^2+\omega^1+\omega^0+\,\dots\,+\omega^{-97}$

$\small\longrightarrow S=\left(\omega^2+\omega^1+\omega^0\right)+\left(\omega^{-1}+\omega^{-2}+\omega^{-3}\right)+\,\dots\,+\left(\omega^{-94}+\omega^{-95}+\omega^{-96}\right)+\omega^{-97}$

$\longrightarrow S=\left(\omega^2+\omega+1\right)+\omega^{-3}\left(\omega^2+\omega+1\right)+\,\dots\,+\omega^{-96}\left(\omega^2+\omega+1\right)+\omega^{-97}$

$\small\longrightarrow S=\left(\omega^2+\omega+1\right)\left(1+\omega^{-3}+\omega^{-6}+\,\dots\,+\omega^{-96}\right)+\omega^{-97}$

Since $\omega^2+\omega+1=0,$

$\longrightarrow S=\omega^{-97}$

$\longrightarrow S=\omega^{-99+2}$

$\longrightarrow S=\omega^{-99}\cdot\omega^2$

$\longrightarrow S=\left(\omega^3\right)^{-33}\cdot\omega^2$

Since $\omega^3=1,$

$\longrightarrow S=\left(1\right)^{-33}\cdot\omega^2$

$\longrightarrow\underline{\underline{S=\omega^2}}$

Answer 2

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \boxed{ \blue{ \bf \: 1 +w + {w}^{2} \: = 0}}$

$(2). \: \boxed{ \blue{ \bf \: {w}^{3} = \: 1}}$

$\boxed{ \pink{\rm :\implies\:\dfrac{1}{w} = {w}^{2} \: \: \: and \: \: \dfrac{1}{ {w}^{2} } = w}}$

where,

$\rm \bullet\:w \: is \: called \: complex \: cube \: roots \: of \: unity.$

$(3). \: \boxed{ \blue{ \bf \: S_n \: = \: \dfrac{a(1 - {r}^{n}) }{1 - r} }}$

Wʜᴇʀᴇ,

• Sₙ is the sum of first n term of GP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common ratio.

$\large\underline\purple{\bold{Solution :- }}$

$\rm :  \implies \:\dfrac{1}{w} + \dfrac{1}{ {w}^{2} } + \dfrac{1}{ {w}^{3} } + - - - + \dfrac{1}{ {w}^{100} }$

• This represents a GP series with

$\rm \bullet\: \: first \: term \: (a) \: = \: \dfrac{1}{w} = {w}^{2}$

$\rm \bullet\:common \: ratio \: (r) \: = \dfrac{1}{ {w}^{2} } \div \dfrac{1}{w} = \dfrac{1}{w} = {w}^{2}$

$\rm \bullet\:number \: of \: terms \: (n) = 100$

So,

Now,

$\rm \implies \:\dfrac{1}{w} + \dfrac{1}{ {w}^{2} } + \dfrac{1}{ {w}^{3} } + - - + \dfrac{1}{ {w}^{100} } = \dfrac{ {w}^{2} ( {w}^{100} - 1) }{w - 1}$

$\rm :\implies \:\dfrac{ {w}^{2} ( {w}^{99} \times w - 1)}{w - 1}$

$\rm :\implies \:\dfrac{ {w}^{2} ( { {((w}^{3} )}^{33} \times w - 1)}{w - 1}$

$\rm :\implies \:\dfrac{ {w}^{2} ( {1}^{33} \times w - 1)}{w - 1}$

$\rm :\implies \:\dfrac{ {w}^{2} ( w - 1)}{w - 1}$

$\rm :\implies\: {w}^{2}$

Hence, the value of

$\boxed{ \pink{ \rm :  \implies \:\dfrac{1}{w} + \dfrac{1}{ {w}^{2} } + \dfrac{1}{ {w}^{3} } + - - - + \dfrac{1}{ {w}^{100} } = {w}^{2} }}$