Question

I will fanned and give you a medal !!
1)
Evaluate the summation of 20 times 0.5 to the n minus 1 power, from n equals 3 to 12..

Answer 1

This?

\(\sum\limits_{n=3}^{12}\left(20\cdot 0.5^{n-1}\right)\)

Answer 2

Give me a sec to draw it

Answer 3

\[\sum_{n=3}^{12?} 20(0.5)^{n-1}\]

Answer 4

@tkhunny

Answer 5

Same thing. Well, add them up!

\(20\cdot\left(1/4 + 1/8 + 1/16 + ... + 1/2048\right)\)

What do you get?

Answer 6

Oh sorry, and do i just add straight across?

Answer 7

the summation is a series of sums. The n starts at value 3 and each repetition of the sum the n increases with 1. You repeat the sums until n=12.

That is what tkhunny did for you.

But I'm writing it down into a broken down scheme to help you visualize what you're supposed to do mechanically

\[n=3 \rightarrow 20\times0.5^{3} = ?\]
\[n=4 \rightarrow 20\times0.5^{4} = ?\]
\[n=5 \rightarrow 20\times0.5^{5} = ?\]
...
\[n=12 \rightarrow 20\times0.5^{12} = ?\]

\[\sum_{n=3}^{12}\] then means you add every separate result. So the total sum is the evaluation of the series you posted.

Answer 8

5⋅(1 +1/2+1/4+...+1/512) -- A little easier.

Answer 9

Want me to solve those that you showed me ?

Answer 10

sorry I made a mistake it should be 0.5² when n=3, 0.5^3 when n=4.... etc and when n=12 it should be 0.5^11

Answer 11

Maybe it's a good idea for you to solve it like that once, so you know what you need to do when they ask you to calculate such a summation.

Answer 12

Yes. You should know how to add a finite geometric series. This WILL be required of you on exams.

Answer 13

Give me a sec pretty please

Answer 14

Can you help me please. I am clueless as to what ti do

Answer 15

I Got 1409/2048 but i do not think that is correct

Answer 16

or is it 20.54?

Answer 17

\[20*0.5^{2}+20*0.5^{3}+20*0.5^{4}+20*0.5^{5}+20*0.5^{6}+20*0.5^{7}\]
\[+20*0.5^{8}+20*0.5^{9}+20*0.5^{10}+20*0.5^{11}\]
is what you need to calculate.

Answer 18

Give me a sec

Answer 19

How do you fan someone ?

Answer 20

Just look at it and say to yourself, "Blech!! I NEVER want to do that. There MUST be another way."

Indeed, there is: \(5\cdot\left(1 + 1/2 + 1/4 + ... + 1/512 \right) = 5\cdot\dfrac{1 - \dfrac{1}{1024}}{1 - \dfrac{1}{2}}\)

Thus, we have turned 9 additions, 10 multiplications, and 10 exponentiations into 3 divisions, 2 subtractions and one (1) multiplication. Just counting operations, we've come down from 29 to 6. Way, WAY less to go wrong. Way, WAY fewer rounding problems.

Learn how to create that summation on the fly. This is the salient point of this exercise.