Evaluate the summation of 20 times 0.5 to the n minus 1 power, from n equals 3 to 12..
\(\bf {\huge \Sigma}_{{\color{brown}{ n}}=3}^{12}\quad a_1\cdot r^{{\color{brown}{ n}}-1}\implies a_1\left(\cfrac{1-r^{\color{brown}{ n}}}{1-r}\right)\)
can you explain further my level of confusion is insane @jdoe0001
well... \(\bf a_1 = 20\ and \ r=0.5\) \(\bf a_1=\textit{1st term of sequence}\quad r=\textit{common ratio or "multiplier"}\)
okay so what is the first term of sequence??
so... what you have is pretty much... includes the 1st term and the common ratio thus \(\bf {\huge \Sigma}_{{\color{brown}{ n}}=3}^{{\color{blue}{ 12}}}\quad 20\cdot 0.5^{{\color{brown}{ n}}-1}\implies (20\cdot 0.5^{{\color{brown}{ 3}}-1})\left(\cfrac{1-0.5^{\color{brown}{ {\color{blue}{ 12}}}}}{1-0.5}\right)\)
the 1st term... .in the sequence is 20 HOWEVER you're asked to start adding from the n = 3, or 3rd term thus I used \(\bf (20\cdot 0.5^{{\color{brown}{ 3}}-1})\)
okay so what is next
get the summation :)
how ???/
\(\bf {\huge \Sigma}_{{\color{brown}{ n}}=3}^{{\color{blue}{ 12}}}\quad 20\cdot 0.5^{{\color{brown}{ n}}-1} \\ \quad \\ \implies \Large (20\cdot 0.5^{{\color{brown}{ 3}}-1})\left(\cfrac{1-0.5^{\color{brown}{ {\color{blue}{ 12}}}}}{1-0.5}\right)\)
9.99 19.95 29.98 39.96 so how do i get to one of these
well.... what did you get for the one above?
thats the thing idk how to solve it.. im bad at math and my mom cant help me right now... so i have 0 clue what im doing with this
hmm well.. the idea is that you have to calculate the value so....
say for example.... what does \(\bf (20\cdot 0.5^{{\color{brown}{ 3}}-1})\) give you?
1.5
hm..... may want to recheck it
5
so.... 5 yes try now just the numerator of the fracton :)
.9997
yeap and the bottom will be 1-0.5 which is just 0.5 so just do a 5 * 0.9997 / 0.5 :)
9.997 OMG THANK YOU SO MUCHHHHH
yw
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