Sigma notation in expanded form and evaluating the sum

[Louise Johnson]

Hello can anyone let me know how I did on these? (thats if you can read my notation)

Thank you
sincerly
Louise

Question #1 Write the below in expanded form and evaluate the sum. Answer correct to four decimal places.


\(\displaystyle \L \sum_{k=3}^{11} 2\(1/3) k-1\)

My Answer: a=2(1/3)^3-1 r=2

( i will just include the first and last of the series.)

2(1/3)^3-1+......+.....+......+.....+......+.....+2(1/3)^10-1

Sum9=2(1/3)^2(1-(1/3)^9)/ (1-1/3)
Sum9=.3333

 

[pka]

Why do you post the odd unreadable way that you do?
Are you really serious about mathematics?
If you are the you would learn to use TeX.
At the top of the this page, 'FORUM HELP' can show you how?

 

[Louise Johnson]

Thank you PKA. I knew there had to be a better way! You are the first person to help clue me in.
Thank you again
Sincerly Louise
Ps yes I am serious about mathmatics...
Pss Looks like you could use some work on your grammar?

 

[pka]

Will wonders never cease?

 

[soroban]

Hello, Louise!

There's no need to shout . . .

1) Write the sum in expanded form and evaluate the sum to four decimal places.

. . \(\displaystyle \L \sum_{k=3}^{11} 2\left(\frac{1}{3}\right)^{k-1}\)

Expanded form: \(\displaystyle \L\:S\;=\;2\left(\frac{1}{3}\right)^2\,+\,2\left(\frac{1}{3}\right)^3\,+\,2\left(\frac{1}{3}\right)^4\,+\,\cdots\,+\,2\left(\frac{1}{3}\right)^{10}\)


We have: \(\displaystyle \L\:S \;=\;2\left(\frac{1}{3}\right)^2\,\underbrace{\left[1\,+\,\frac{1}{3}\,+\,\left(\frac{1}{3}\right)^2 \,+\,\left(\frac{1}{3}\right)^3\,+\,\cdots\,\left(\frac{1}{3}\right)^8\right]}_{\text{geometric series}}\)

The series has first term \(\displaystyle a\,=\,1\), common ratio \(\displaystyle r\,=\,\frac{1}{3}\), and \(\displaystyle 9\) terms.

. . Its sum is: \(\displaystyle \L\:1\cdot\frac{1\,-\,(\frac{1}{3})^9}{1\,-\,\frac{1}{3}} \;=\;\frac{\frac{19682}{19683}}{\frac{2}{3}} \;=\;\frac{9841}{6561}\)


Therefore: \(\displaystyle \L\:S\;=\;2\cdot\frac{1}{9}\cdot\frac{9841}{6561} \:=\:\frac{19,682}{59,049} \:\approx\:0.3333\)

 

[Louise Johnson]

Thank you Soroban, That is great to know I have something correct for a change. I have to say you make it look much much better and I am looking forward to the day when I can post it exactly like that !. I am on my way to learning the tex code but its an even slower process than the math!!
Thank you again
Louise