Asked by COFFEE
infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n)
.. my work so far. i used the ratio test
= lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] |
.. now my question is: was it ok for me to add "+1" to "n+2" to become "n+3"?
= lim (n-->infinity) | [((n+3)/(10^(n+1)))*(((x-5)^(n+1))/1)] * [((10^n)/(n+2))*(1/((x-5)^n))] |
= lim (n-->infinity) | [(((n+3)(x-5))/10)*(1/(n+2))] |
how do i finish this so that i could find the endpoints? please help. thank you.
.. my work so far. i used the ratio test
= lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] |
.. now my question is: was it ok for me to add "+1" to "n+2" to become "n+3"?
= lim (n-->infinity) | [((n+3)/(10^(n+1)))*(((x-5)^(n+1))/1)] * [((10^n)/(n+2))*(1/((x-5)^n))] |
= lim (n-->infinity) | [(((n+3)(x-5))/10)*(1/(n+2))] |
how do i finish this so that i could find the endpoints? please help. thank you.
Answers
Answered by
roger
what is the answer x-5/3=10
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