Expert: Ahmed Salami - 7/28/2010

Question
QUESTION: I don't understand how to do these sigmas? or whatever they're called? I attached an image with the questions.. If you could please help me out.

ANSWER: Hi Victoria,
You simply need to evaluate the expression for every number in the range and sum them up.
For example, in the first question you're required to find all the values of 3k from k = 1 to k = 8 and add them up.
At k = 1, 3k = 3(1) = 3
At k = 2, 3k = 3(2) = 6
At k = 3, 3k = 3(3) = 9
.
.
At k = 8, 3k = 3(8) = 24
You get the idea.
Now you can do the rest. And you can always get back to me.

Regards

---------- FOLLOW-UP ----------

QUESTION: What about for number two? The.. 4^n+2

Sorry, I'm not very good in math..

Answer
Hi Victoria,
Its ok, you'll get better.
At n = 1, 4^(n+2) = 4^(1+2) = 4^3 = (4)(4)(4) = 64
At n = 2, 4^(n+2) = 4^(2+2) = 4^4 = (4)(4)(4)(4) = 256
At n = 3, 4^(n+2) = 4^(3+2) = 4^5 = (4)(4)(4)(4)(4) = 1024
At n = 4, 4^(n+2) = 4^(4+2) = 4^6 = (4)(4)(4)(4)(4)(4) = 4096

Also, for 4^(-k)
At k = 0, 4^(-k) = 4^(0) = 1/(4^0) = 1/1 = 1
At k = 1, 4^(-k) = 4^(-1) = 1/(4^1) = 1/4 = 0.25
At k = 2, 4^(-k) = 4^(-2) = 1/(4^2) = 1/16 = 0.0625
At k = 3, 4^(-k) = 4^(-3) = 1/(4^3) = 1/64 = 0.015625

For (-k)^(k+1),
At k = 1, (-k)^(k+1) = (-1)^(1+1) = (-1)^2 = (-1)(-1) = 1
At k = 2, (-k)^(k+1) = (-2)^(2+1) = (-2)^3 = (-2)(-2)(-2) = -8
At k = 3, (-k)^(k+1) = (-3)^(3+1) = (-3)^4 = (-3)(-3)(-3)(-3) = 81
At k = 4, (-k)^(k+1) = (-4)^(4+1) = (-4)^5 = (-4)(-4)(-4)(-4)(-4) = -1024

Happy to help.

Regards