Question #d7237

1 Answer
Mar 21, 2017

Answer:

#color(blue)("Thus the sum must be converging")#
#color(blue)(lim_(n->oo) s=30xx5/3=50)#

#color(blue)(s=sum_(i=1to oo)30(2/5)^(i-1) )#

Explanation:

#a_1=30r^0 ->30r^(1-1)#

#a_2=30r^1->30r^(2-1)#

#a_3=30r^2->30r^(3-1)#

As #r=2/5# then #r^i# is decreasing as #i# is increasing

Thus #30r^i# is decreasing as #i# is increasing

#color(blue)("Thus the sum must be converging")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Set the sum to be s

#s=30+30(2/5)+30(2/5)^2+30(2/5)^3+.. +30(2/5)^n#

#s(2/5)=30(2/5)+30(2/5)^2+..+30(2/5)^n+30(2/5)^(n+1)#

#s(1-2/5)=30-30(2/5)^(n+1) = 30[1-(2/5)^(n+1)]#

#s= 30xx[1-(2/5)^(n+1)]/(1-2/5)#

As this is an infinite series we have

#lim_(n->oo) s=30xx(1-0)/(3/5)#

#color(blue)(lim_(n->oo) s=30xx5/3=50)#

#color(blue)(s=sum_(i=1to oo)30(2/5)^(i-1) )#