Mathematics: The first terms of an infinite geometric sequence, un are 2, 6, 18, the first terms of a second infinite geometric sequence, vn are 2, -6, 18, - the terms of a third sequence wn are defined as wn = un + vn 1. the finite series, summation: upper limit= 225, lower limit k=1, formula wk can also be written in the form summation: upper limit= m, lower limit k=0, formula 4r^k. find the value of r. 2. find the value of m

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The first terms of an infinite geometric sequence,

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cookiebrain72

14.01.2022

Mathematics

r = 9 and m = 112

Step-by-step explanation:

$\sum_{k=1}^{225}W_{k}=\sum_{k=0}^{m}4r^{k}$

Write W in terms of U and V.

$\sum_{k=1}^{225}(U_{k}+V_{k})=\sum_{k=0}^{m}4r^{k}\\\sum_{k=1}^{225}U_{k}+\sum_{k=1}^{225}V_{k}=\sum_{k=0}^{m}4r^{k}$

Define U and V using geometric series formula.

$\sum_{k=1}^{225}2(3)^{k-1}+\sum_{k=1}^{225}2(-3)^{k-1}=\sum_{k=0}^{m}4r^{k}$

Use sum of geometric series formula.

$2(\frac{1-(3)^{225}}{1-3})+2(\frac{1-(-3)^{225}}{1-(-3)})=4(\frac{1-(r)^{m+1}}{1-r})$

Simplify.

$-1(1-3^{225})+\frac{1+3^{225}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\-1+3^{225}+\frac{1}{2}+\frac{3^{225}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\-\frac{1}{2}+\frac{3(3^{225})}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\\frac{-1+3(3^{225})}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\\frac{-1+3^{226}}{2}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{-1+3^{226}}{8}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{1-3^{226}}{-8}=4(\frac{1-(r)^{m+1}}{1-r})\\4\frac{1-9^{113}}{1-9}=4(\frac{1-(r)^{m+1}}{1-r})$

Therefore, r = 9 and m = 112.

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