Imaginary summation?

Algebra Level 5

If $n$ is an integer and $\omega$ is an imaginary cube root of unity, let $A$ denote the value of

${\displaystyle\sum_{k=0}^{n} \binom{n}{k} \omega^{k}}$

Then the number of possible values of $e^A$ is?

Note :This question is a part of set KVPY 2014 SB

2

3

6

4

2 solutions

Harshith Reddy
Feb 19, 2015

i have an easier solution 1+w+w^2=0 so 1+W=-w^2 so the problem is reduced to number of vlues of (-w2)^n one can easily see tht it can take the values 1,-1,w,w2,-w,-w2 for different values of n

Mahimn Bhatt
Feb 13, 2015

Don't mind but it is overrated question.

Apply binomial theorem, (1+x) ^ n

Putting x=w

For n=1,2,3,4,5,6 we get different values and for n=7 the cycle is repeated.! Hence 6 different values.

Well, n is any integer

Kartik Sharma - 6 years, 3 months ago

What if ω=1? Then there can be infinite values

Archisman Bhattacharjee - 4 years, 7 months ago

Note that the problem states that $\omega$ is an imaginary cube root of unity, so you can't have $\omega=1$ .

Prasun Biswas - 4 years, 7 months ago

I agree! It was level 3 or 4 in beginning.

Pranjal Jain - 6 years, 4 months ago

Imaginary summation?

2 solutions

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