Question 172793
Take note that the terms form a geometric sequence (since there is a common ratio and exponent involved)



Recall, the formula for a geometric sequence is {{{a[n]=a*r^n}}} where {{{n>=1}}}



Since the coefficient is 300, this tells us that {{{a=300}}}



Since the value being raised to a power is {{{-4/5}}}, this means that {{{r=-4/5}}}



So the formula is {{{a[n]=300(-4/5)^n}}}. So, for instance, if {{{n=3}}}, then {{{a[3]=300(-4/5)^3}}} (which is the third term).



Also, remember that the formula for an infinite geometric series is



{{{sum(a*r^k,k=0,infinity)=a/(1-r)}}}


Since we're starting at n=1, the series needs to be rewritten to


{{{sum(a*r^k,k=1,infinity)=a/(1-r)-a}}}




So in this case, the formula we'll use is 


{{{sum(300*(-4/5)^n,n=1,infinity)=300/(1-(-4/5))-300=300/(1+4/5)-300=300/(9/5)-300=500/3-300=-400/3}}}




So the sum of the infinite series is {{{-400/3}}}