calculation of sum, (1/6)*( sum_{n=-2 to 3} (1/2)* e^{-j (pi/3) k} )

[avishaib]

Hi can someone explain me this?

---

\(\displaystyle \dfrac{1}{6}\,\)\(\displaystyle \displaystyle \sum_{n\, =\, -2}^3\, \)\(\displaystyle \left(\, \dfrac{1}{2}\, e^{-j\, \dfrac{\pi}{3}\, k}\,\right)^n\, =\, \dfrac{\dfrac{1}{2^5}\, -\, 2}{1.5\, \cdot\, (-1)^k\, -\, 3\, \cdot\, e^{-j\, \dfrac{\pi}{3}\, k}}\)

---

Thank you

 

[stapel]

Hi can someone explain me this?

---

\(\displaystyle \dfrac{1}{6}\,\)\(\displaystyle \displaystyle \sum_{n\, =\, -2}^3\, \)\(\displaystyle \left(\, \dfrac{1}{2}\, e^{-j\, \dfrac{\pi}{3}\, k}\,\right)^n\, =\, \dfrac{\dfrac{1}{2^5}\, -\, 2}{1.5\, \cdot\, (-1)^k\, -\, 3\, \cdot\, e^{-j\, \dfrac{\pi}{3}\, k}}\)

The left-hand side is six terms. What did you get when you expanded the expression on the left-hand side, converted everything to a common denominator, and then simplified into one fractional expression?

Please be complete, including what summation formulas you applied. Thank you!

 

[pka]

Hi can someone explain me this?

---

\(\displaystyle \dfrac{1}{6}\,\)\(\displaystyle \displaystyle \sum_{n\, =\, -2}^3\, \)\(\displaystyle \left(\, \dfrac{1}{2}\, e^{-j\, \dfrac{\pi}{3}\, k}\,\right)^n\, =\, \dfrac{\dfrac{1}{2^5}\, -\, 2}{1.5\, \cdot\, (-1)^k\, -\, 3\, \cdot\, e^{-j\, \dfrac{\pi}{3}\, k}}\)

You need to explain some notation to us. Do we assume that \(\displaystyle j^2=-1\) rather than what mathematicians use which is \(\displaystyle i\).
If so then you need to know that \(\displaystyle e^{ix}=\cos(x)+i\sin(x)\).
Also you told us nothing about what \(\displaystyle k\) might be.

 

[Ishuda]

Hi can someone explain me this?

---

\(\displaystyle \dfrac{1}{6}\,\)\(\displaystyle \displaystyle \sum_{n\, =\, -2}^3\, \)\(\displaystyle \left(\, \dfrac{1}{2}\, e^{-j\, \dfrac{\pi}{3}\, k}\,\right)^n\, =\, \dfrac{\dfrac{1}{2^5}\, -\, 2}{1.5\, \cdot\, (-1)^k\, -\, 3\, \cdot\, e^{-j\, \dfrac{\pi}{3}\, k}}\)

---

Thank you

Let
x = \(\displaystyle \dfrac{1}{2}\, e^{-j\, \dfrac{\pi}{3}\, k}\)
and S be the sum. Then you have
S = \(\displaystyle \dfrac{1}{6\, x^2} \underset{n=0}{\overset{n=5}{\Sigma}}\, x^n\)
or
S = \(\displaystyle \dfrac{1}{6\, x^2} \dfrac{1\, -\, x^6}{1\, -\, x}\)

Now simplify, i.e
x⁶ = \(\displaystyle \dfrac{1}{2^6}\, [ cos(\dfrac{6\, k\, \pi}{3})\, -\, j\, sin(\dfrac{6\, k\, \pi}{3})\, ]\)
= \(\displaystyle \dfrac{1}{2^6}\)
etc. ..