If w is an imaginary cube root of unity, then find value of expression
2(1 + w) (1 + w^2) +3 (2 + w) (2 + w^2) + …+ (n + 1) (n + w) (n + w^2) in terms of n.
(k+w)(k+w^2) = w^3+kw^2+k^2w+k^3 = = (w^4-k^4)/(w-k)
See where this takes you:
https://www.wolframalpha.com/input/?i=sum(k%3D1..n)+(w%5E4-k%5E4)%2F(w-k)
To find the value of the given expression in terms of n, we need to simplify it step by step. Let's break it down into smaller parts.
First, let's consider the expression (1 + w) (1 + w^2). We can expand this using the distributive property:
(1 + w) (1 + w^2) = 1*1 + 1*w^2 + w*1 + w*w^2
= 1 + w^2 + w + w^3
Note that w is an imaginary cube root of unity, which means w^3 = 1. So our expression simplifies to:
1 + w^2 + w + 1
= 2 + w + w^2
Similarly, let's consider the expression (2 + w) (2 + w^2):
(2 + w) (2 + w^2) = 2*2 + 2*w^2 + w*2 + w*w^2
= 4 + 2w^2 + 2w + 2w^3
(Note: We replaced w^3 with 1 since w is an imaginary cube root of unity)
Again, simplifying further:
4 + 2w^2 + 2w + 2
= 6 + 2w^2 + 2w
Now, let's generalize this pattern to the rest of the terms in the given expression:
(n + w) (n + w^2) = n*n + n*w^2 + w*n + w*w^2
= n^2 + nw^2 + wn + w^3
(Note: We replaced w^3 with 1 since w is an imaginary cube root of unity)
Simplifying further:
n^2 + nw^2 + wn + 1
Now, we can express the entire given expression in terms of n:
2(1 + w) (1 + w^2) + 3 (2 + w) (2 + w^2) + … + (n + 1) (n + w) (n + w^2)
= 2(2 + w + w^2) + 3(6 + 2w^2 + 2w) + … + (n + 1)(n^2 + nw^2 + wn + 1)
Expanding further by distributing:
= 4 + 2w + 2w^2 + 18 + 6w^2 + 6w + … + (n^3 + n^2w^2 + n^2w + nw + n^2 + nw^2 + wn + w + n + 1)
Now, let's group the terms with similar powers of w:
= (4 + 18 + … + n^3) + (2w + 6w + … + nw) + (2w^2 + 6w^2 + … + n^2w^2) + … + (2 + 6 + … + n)
The first group is a sum of consecutive cubes: 4 + 18 + … + n^3, which can be simplified as:
= (1^3 + 2^3 + … + n^3) = (sum of cubes from 1 to n)
The second group is a sum of arithmetic progressions with a common difference of w:
= 2w + 6w + … + nw = w(2 + 6 + … + n) = w(sum of numbers from 2 to n)
Similarly, we can simplify the third and subsequent groups:
= 2w^2(1 + 3 + … + n-1)
= 2(1 + 3 + … + n-1) (since w^2 = -w - 1)
Finally, we can write the expression in terms of n:
(sum of cubes from 1 to n) + w(sum of numbers from 2 to n) + 2(sum of odd numbers from 1 to n-1)
So, the value of the given expression in terms of n is:
(sum of cubes from 1 to n) + w(sum of numbers from 2 to n) + 2(sum of odd numbers from 1 to n-1)