How do you solve #4.5xx10^9 = -1.8xx10^10 + lim_(N->40) sum_(n=1)^(N) x/(1.04n)#?

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Answer 1 · 2015-09-26T01:11:59

It kind of looks weird, but I'll see how it turns out I guess...

#4.5xx10^9 + 1.8xx10^10 = lim_(N->40) sum_(n=1)^N x/(1.04n)#

#x# of course multiplies by everything, so you can factor it out at the end to get:

#2.25xx10^10 = x sum_(n=1)^(40) 1/(1.04n)#

#x = (2.25xx10^10) / (sum_(n=1)^(40) 1/(1.04n))#

If you bother to do the addition on the bottom, you get:

#x = (2.25xx10^10) / (4.11398)#

#= color(blue)(5.469xx10^9)#