How do you find the sum of the finite geometric sequence of #Sigma 2(-1/4)^n# from n=0 to 40?

Question

Aurora Alvarado · Accepted Answer

#(8/5)*(1+(1/4)^41)#
#~=1.60000000...# There are 41 terms in the series. The common ratio, #r# is #-(1/4)# and the first term, #a# is #2(-1/4)60# which is #2#. Therefore the sum of the first 41 terms is #a*(1-r^n)/(1-r)# #=2*(1-(-1/4)^41)/(1-(-1/4))# #=8/5*(1+(1/4)^41)# because 41 is odd #~=1.6# because #(1/4)^41# is negligible

Charles Arocha · Answer

undefined The sum of the finite geometric sequence Σ 2(-1/4)^n from n=0 to 40 can be found using the formula for the sum of a finite geometric series: $S = a \frac{{1 - r^{n+1}}}{{1 - r}}$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. In this case, $a = 2$, $r = -1/4$, and $n = 40$. Substituting these values into the formula yields:

$S = 2 \frac{{1 - (-1/4)^{40+1}}}{{1 - (-1/4)}}$

$S = 2 \frac{{1 - (-1/4)^{41}}}{{1 + 1/4}}$

$S = 2 \frac{{1 - (-1/4)^{41}}}{{5/4}}$

$S = \frac{{8}}{{5}}(1 - (-1/4)^{41})$

$S ≈ \frac{{8}}{{5}}$

So, the sum of the finite geometric sequence Σ 2(-1/4)^n from n=0 to 40 is approximately $\frac{{8}}{{5}}$.