Complex Numbers And Quadratic Equations question 159
Question: The value of the expression $ 1.(2-\omega )(2-{{\omega }^{2}})+2.(3-\omega )(3-{{\omega }^{2}})+……. $ $ ….+(n-1).(n-\omega )(n-{{\omega }^{2}}), $ where $ \omega $ is an imaginary cube root of unity, is [IIT 1996]
Options:
A) $ \frac{1}{2}(n-1)n(n^{2}+3n+4) $
B) $ \frac{1}{4}(n-1)n(n^{2}+3n+4) $
C) $ \frac{1}{2}(n+1)n(n^{2}+3n+4) $
D) $ \frac{1}{4}(n+1)n(n^{2}+3n+4) $
Show Answer
Answer:
Correct Answer: B
Solution:
$ r^{th} $ term of the given series = $ r[(r+1)-\omega ][(r+1)-{{\omega }^{2}}] $ = $ r[{{(r+1)}^{2}}-(\omega +{{\omega }^{2}})(r+1)+{{\omega }^{3}}] $ = $ r[{{(r+1)}^{2}}-(-1)(r+1)+1] $ = $ r[(r^{2}+3r+3]=r^{3}+3r^{2}+3r $ Thus sum of the given series $ =\sum\limits_{r=1}^{(n-1)}{(r^{3}+3r^{2}+3r)} $ $ =\frac{1}{4}{{(n-1)}^{2}}n^{2}+3.\frac{1}{6}(n-1)(n)(2n-1)+3.\frac{1}{2}(n-1)n $ $ =\frac{1}{4}(n-1)n(n^{2}+3n+4) $
BETA
