Question

Sigma notation help? Will give medal

Answer 1

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

Answer 2

hint:
I can rewrite your sum as follows:
\[\begin{gathered}
40\sum\limits_{n = 3}^{12} {\frac{1}{{{2^n}}}} = \hfill \\
= 40\left( {\frac{1}{{{2^3}}} + \frac{1}{{{2^4}}} + \frac{1}{{{2^5}}} + \frac{1}{{26}} + \frac{1}{{{2^7}}} + \frac{1}{{{2^8}}} + \frac{1}{{{2^9}}} + \frac{1}{{{2^{10}}}} + \frac{1}{{{2^{11}}}} + \frac{1}{{{2^{12}}}}} \right) = ...? \hfill \\
\end{gathered} \]

Answer 3

Ohhh okay I was plugging it in wrong. The lesson doesn't give much info on how to solve these so I've been lost..

Answer 4

\[\begin{gathered}
40\sum\limits_{n = 3}^{12} {\frac{1}{{{2^n}}}} = \hfill \\
= 40\left( {\frac{1}{{{2^3}}} + \frac{1}{{{2^4}}} + \frac{1}{{{2^5}}} + \frac{1}{{{2^6}}} + \frac{1}{{{2^7}}} + \frac{1}{{{2^8}}} + \frac{1}{{{2^9}}} + \frac{1}{{{2^{10}}}} + \frac{1}{{{2^{11}}}} + \frac{1}{{{2^{12}}}}} \right) = \hfill \\
= \frac{{40}}{8}\left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{2^4}}} + \frac{1}{{{2^5}}} + \frac{1}{{{2^6}}} + \frac{1}{{{2^7}}} + \frac{1}{{{2^8}}} + \frac{1}{{{2^9}}}} \right) = ...? \hfill \\
\end{gathered} \]

Answer 5

so we have to compute the sum into the parentheses

Answer 6

So if I round it I'll get 9.99? (btw that's one of the answers it gives me)

Answer 7

Please wait I'm checking your answer...

Answer 8

When you are done could you help me with one more? I'm not getting this one no matter how I try to solve it
\[\sum_{n=1}^{12}2n+5\]

Answer 9

ok!

Answer 10

Thank you :)

Answer 11

your first result is right!

Answer 12

I got the answer 216 for \[\sum_{n=1}^{12}2n+5\]
But I'm not sure if that's correct. And thank you :)

Answer 13

here, we can rewrite your sum, as follows:
\[\large \left( {\sum\limits_{n = 1}^{12} {2n} } \right) + 5 \times 12 = \left( {\sum\limits_{n = 1}^{12} {2n} } \right) + 60 = ...?\]

Answer 14

and:
\[\large \begin{gathered}
\left( {\sum\limits_{n = 1}^{12} {2n} } \right) + 5 \times 12 = \left( {\sum\limits_{n = 1}^{12} {2n} } \right) + 60 = 2\left( {\sum\limits_{n = 1}^{12} n } \right) + 60 = \hfill \\
\hfill \\
= 2\frac{{1 + 12}}{2} \times 12 + 60 = 13 \times 12 + 60 = 216 \hfill \\
\end{gathered} \]

Answer 15

so your answer is right!

Answer 16

Thank you so much for your help :)

Answer 17

Thank you! :)