Question

The value of the expression $1.(2-w)(2-w^2)+2.(3-w)(3-w^2)+..........+(n-1)(n-w)(n-w^2),$
where ω is an imaginary cube root of unity , is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

CONVENTIONAL APPROACH
rth term of the given series
= $r\left[\right(r+1)-w]\left[\right(r+1)-w^2]$

=$r\left[\right(r+1{)}^2-(w+w^2)(r+1)+w^3]$
=$r\left[\right(r+1{)}^2-(-1)(r+1)+1]$
$r\left[\right(r^2+3r+3]=r^2+3r^2+3r$

Thus sum of the given series $=\sum _{r=1}^{(n-1)}(r^3+3r^2+3r)$

$=\frac{1}{4}(n-1{)}^2{n}^2+3.\frac{1}{6}(n-1\left)\right(n\left)\right(2n-1)+3.\frac{1}{2}(n-1)n$
$=\frac{1}{4}(n-1)n(n^2+3n+4)$

This is avariable to variable question i.e. the question is in variable and the options are in variables. So, we can assume and substitute any value for the variables.

Tricks: Put n=2,then the first term $=1.(2-w)(2-w^2)$
$=4-2(w+w^2)+w^3$
$=4-2(-1)+1$
$=7$
Now, put n=2 in all the answer options and eliminate the options where the answer is not equal to 7.
Hence,only(b) remains.