Question

If $\omega$ is an imaginary cube root of unity, then the value of $2\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right) + 3\left( {2 + \omega } \right)\left( {2 + {\omega ^2}} \right) + ...... + 101\left( {100 + \omega } \right)\left( {100 + {\omega ^2}} \right)$ is-

Answer 1

If $\omega$is the cube root of unity then $1 + \omega + {\omega ^2} = 0$ and ${\omega ^3} = 1$. Take the first term of the series and solve it using these identities then repeat the same process with the second term and last term. Substitute the values of all the terms in $2\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right) + 3\left( {2 + \omega } \right)\left( {2 + {\omega ^2}} \right) + ...... + 101\left( {100 + \omega } \right)\left( {100 + {\omega ^2}} \right)$.You will see a common pattern emerging so you write it in summation form. Solve the summation to get the answer.

Complete step-by-step answer:
If $\omega$is the cube root of unity then it satisfies two properties-
${\omega ^3} = 1$ and $1 + \omega + {\omega ^2} = 0$
Now we have to find the value of $2\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right) + 3\left( {2 + \omega } \right)\left( {2 + {\omega ^2}} \right) + ...... + 101\left( {100 + \omega } \right)\left( {100 + {\omega ^2}} \right)$-- (i)
Let us first take first term of the series-$2\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right)$
On solving this we get,
$\Rightarrow 2\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right) = 2\left( {1 + {\omega ^2} + \omega + {\omega ^3}} \right)$
On rearranging we get-
$\Rightarrow 2\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right) = 2\left( {1 + {\omega ^3} + {\omega ^2} + \omega } \right)$
Now we know that $1 + \omega + {\omega ^2} = 0 \Rightarrow \omega + {\omega ^2} = - 1$
On using this value in above equation we get,
$\Rightarrow 2\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right) = 2\left( {1 + {\omega ^3} - 1} \right)$
And we can write $2 = \left( {1 + 1} \right)$ then the equation becomes,
$\Rightarrow 2\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right) = \left( {1 + 1} \right)\left( {1 + {\omega ^3} - 1} \right)$ --- (ii)
On taking second term of the series,
$\Rightarrow 3\left( {2 + \omega } \right)\left( {2 + {\omega ^2}} \right) = 3\left( {{2^2} + {\omega ^3} + 2{\omega ^2} + 2\omega } \right)$
On taking $2$ common we get,
$\Rightarrow 3\left( {2 + \omega } \right)\left( {2 + {\omega ^2}} \right) = 3\left( {{2^2} + {\omega ^3} + 2\left( {{\omega ^2} + \omega } \right)} \right)$
Again we know that $1 + \omega + {\omega ^2} = 0 \Rightarrow \omega + {\omega ^2} = - 1$
On using this in the formula we get,
$\Rightarrow 3\left( {2 + \omega } \right)\left( {2 + {\omega ^2}} \right) = 3\left( {{2^2} + {\omega ^3} + 2\left( { - 1} \right)} \right)$
On simplifying we get,
$\Rightarrow 3\left( {2 + \omega } \right)\left( {2 + {\omega ^2}} \right) = 3\left( {{2^2} + {\omega ^3} - 2} \right)$
Now we can write $3 = \left( {2 + 1} \right)$
On putting this value in above equation we get,
$\Rightarrow 3\left( {2 + \omega } \right)\left( {2 + {\omega ^2}} \right) = \left( {2 + 1} \right)\left( {{2^2} + {\omega ^3} - 2} \right)$ --- (iii)
Similarly on taking last term of the series and solving it in same method we get,
$\Rightarrow 101\left( {100 + \omega } \right)\left( {100 + {\omega ^2}} \right) = \left( {100 + 1} \right)\left( {{{100}^2} + {\omega ^3} - 100} \right)$ -- (iv)
On putting values of eq. (ii), (iii) and (iv) in eq. (i), we get-
$\Rightarrow \left( {1 + 1} \right)\left( {{1^2} + {\omega ^3} - 1} \right) + \left( {2 + 1} \right)\left( {{2^2} + {\omega ^3} - 2} \right) + ...... + \left( {100 + 1} \right)\left( {{{100}^2} + {\omega ^3} - 100} \right)$
On observing this series, we see that this series is in the form of
$\Rightarrow$ $\sum\limits_{i = 1}^n {\left( {n + 1} \right)} \left( {{n^2} + {\omega ^3} - n} \right)$
Where n=$100$
So we can write the series as-
$\Rightarrow \sum\limits_{i = 1}^{100} {\left( {n + 1} \right)} \left( {{n^2} + {\omega ^3} - n} \right)$
Now we know that ${\omega ^3} = 1$
So on putting this value we get,
$\Rightarrow \sum\limits_{i = 1}^{100} {\left( {n + 1} \right)} \left( {{n^2} + 1 - n} \right)$
On solving multiplying $\left( {n + 1} \right)$ inside the bracket, we get
$\Rightarrow \sum\limits_{i = 1}^{100} {\left( {n\left( {{n^2} + 1 - n} \right) + 1\left( {{n^2} + 1 - n} \right)} \right)}$
On multiplication we get,
$\Rightarrow \sum\limits_{i = 1}^{100} {\left( {{n^3} + n - {n^2} + {n^2} + 1 - n} \right)}$
On cancelling common terms with opposite sign we get,
$\Rightarrow \sum\limits_{i = 1}^{100} {\left( {{n^3} + 1} \right)}$
$\Rightarrow \sum\limits_{i = 1}^{100} {{n^3}} + \sum\limits_{i = 1}^{100} 1$
Now we know that $\sum\limits_{i = 1}^n {{m^3}} = {\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2}$ where m is a natural number
And $\Rightarrow \sum\limits_{i = 1}^n {1 = n}$
On applying the values in the formula we get,
$\Rightarrow {\left( {\dfrac{{100\left( {100 + 1} \right)}}{2}} \right)^2} + 100$
On solving we get,
$\Rightarrow {\left( {\dfrac{{100 \times 101}}{2}} \right)^2} + 100$
$\Rightarrow {\left( {50 \times 101} \right)^2} + 100$
On multiplication we get,
$\Rightarrow {\left( {5050} \right)^2} + 100$
On squaring the given number we get,
$\Rightarrow 25502500 + 100$
On adding we get,
$\Rightarrow 25502600$
Hence the correct answer is $25502600$.

Note: Here we can also use the formula for the series in the form -
$\Rightarrow \sum\limits_{i = 1}^n {\left( {n + 1} \right)} \left( {{n^2} + 1 - n} \right)$
The formula for the sum of such form of series is-
$\Rightarrow {S_n} = \dfrac{{{n^2}}}{4}{\left( {n + 1} \right)^2} + n$
So on putting n=$100$ we get,
$\Rightarrow {S_n} = \dfrac{{{{100}^2}}}{4}{\left( {100 + 1} \right)^2} + 100$
$\Rightarrow {S_n} = \dfrac{{10000}}{4} \times {101^2} + 100$
$\Rightarrow {S_n} = \left( {2500 \times 10201} \right) + 100$
$\Rightarrow {S_n} = 25502600$
Hence we get the answer.