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Q. The value of the expression `2(1+1/omega)(1+1/(omega)^2) +...

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user8235304 | Student, Grade 11 | Valedictorian

Posted August 18, 2013 at 11:26 AM via web

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Q. The value of the expression `2(1+1/omega)(1+1/(omega)^2) + 3(2+1/omega)(2+1/(omega)^2) + 4(3+1/omega)(3+1/(omega)^2)...............+(n+1)(n+1/omega)(n+1/(omega)^2)` ` `

` ` where `omega` is an imaginary cube root of unity is

A) `[n(n^2+2)]/3`

B) `[n(n^2-2)]/3`

C) `[n^2(n+1)^2 + 4n]/4`

D) none of these

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 18, 2013 at 2:46 PM (Answer #1)

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You need to evaluate the value of the expression `2(1 + 1/omega)(1 + 1/omega^2) + 3(2 + 1/omega)(2 + 1/omega^2) + 4(3 + 1/omega)(3 + 1/omega^2) + ... +(n+1)(n + 1/omega)(n + 1/omega^2)` .

The problem provides the information that omega is the imaginary cube root of unity, hence, omega checks the equation `omega^2 + omega + 1 = 0` .

Performing the multiplications yields:

`2(1 + 1/omega + 1/omega^2 + 1/omega^3) + 3(2^2 + 2(1/omega + 1/omega^2) + 1/omega^3) +.... + (n+1)(n^2 + n(1/omega + 1/omega^2) + 1/omega^3)`

Since `omega^3 = 1` yields:

`2(1 + 1/omega + 1/omega^2 + 1) + 3(2^2 + 2(1/omega + 1/omega^2) + 1) +.... + (n+1)(n^2 + n(1/omega + 1/omega^2) + 1)`

You need to evaluate the summation `1/omega + 1/omega^2` such that:

`1/omega + 1/omega^2 =(omega + 1)/omega^2`

Using the equation `omega^2 + omega + 1 = 0` yields `omega + 1 = -omega^2` , hence, you need to replace `-omega^2` for `omega + 1 ` such that:

`1/omega + 1/omega^2 = -omega^2/omega^2 = -1`

`2(1^2 - 1 + 1) + 3(2^2 - 2 + 1) + 4(3^2 - 3 + 1).... + (n+1)(n^2 - n + 1) = 2*1^2 + 3*2^2 + 4*3^2 + ... + (n+1)*n^2 + 2*(-1) + 3*(-2) + 4*(-3) + ... + (n+1)*(-n) + 2 + 3 + 4 + .... + (n+1)`

`2(1^2 - 1 + 1) + 3(2^2 - 2 + 1) + 4(3^2 - 3 + 1).... + (n+1)(n^2 - n + 1) = sum_(k=1)^n (k+1)*k^2 - sum_(k=1)^n (k+1)*k + ((2 + n+1)*n)/2`

`2(1^2 - 1 + 1) + 3(2^2 - 2 + 1) + 4(3^2 - 3 + 1).... + (n+1)(n^2 - n + 1) = sum_(k=1)^n(k^3 + k^2) - sum_(k=1)^n (k^2 + k) + (n(n+3))/2`

`2(1^2 - 1 + 1) + 3(2^2 - 2 + 1) + 4(3^2 - 3 + 1).... + (n+1)(n^2 - n + 1) = sum_(k=1)^n k^3 + sum_(k=1)^n k^2 - sum_(k=1)^n k^2 - sum_(k=1)^n k + (n(n+3))/2`

Reducing duplicate members yields:

`2(1^2 - 1 + 1) + 3(2^2 - 2 + 1) + 4(3^2 - 3 + 1).... + (n+1)(n^2 - n + 1) = sum_(k=1)^n k^3 - sum_(k=1)^n k + (n(n+3))/2`

`2(1^2 - 1 + 1) + 3(2^2 - 2 + 1) + 4(3^2 - 3 + 1).... + (n+1)(n^2 - n + 1) = (n(n+1))^2/4 - (n(n+1))/2+ (n(n+3))/2 `

`2(1^2 - 1 + 1) + 3(2^2 - 2 + 1) + 4(3^2 - 3 + 1).... + (n+1)(n^2 - n + 1) = (n(n+1))^2/4 + (n(n + 3 - n - 1))/2`

`2(1^2 - 1 + 1) + 3(2^2 - 2 + 1) + 4(3^2 - 3 + 1).... + (n+1)(n^2 - n + 1) = (n(n+1))^2/4 + n`

Bringing the terms to a common denominator, yields:

`2(1^2 - 1 + 1) + 3(2^2 - 2 + 1) + 4(3^2 - 3 + 1).... + (n+1)(n^2 - n + 1) = (n^2(n+1)^2 + 4n)/4`

Hence, evaluating the given expression, under the given conditions, yields the valid answer `C) (n^2(n+1)^2 + 4n)/4.`

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