How do you find the sum of #Sigma (3^n+4^n)/5^n# from n is #[0,oo)#?

1 Answer
Steve M
Feb 18, 2017

# sum_(n=0)^oo (3^n+4^n)/5^n = 15/2 #

Explanation:

# sum_(n=0)^oo (3^n+4^n)/5^n=sum_(n=0)^oo (3^n/5^n + 4^n/5^n) #
# " "= sum_(n=0)^oo 3^n/5^n + sum_(n=0)^oo 4^n/5^n #
# " "= sum_(n=0)^oo (3/5)^n + sum_(n=0)^oo (4/5)^n #

The 1st sum is a GP with 1st term #a=1#, and common ratio #r=3/5#
The 2nd sum is a GP with 1st term #a=1#, and common ratio #r=4/5#

Using the GP formula: #S_n=a/(1-r)#, we get:

# sum_(n=0)^oo (3^n+4^n)/5^n = (1/(1-3/5)) + (1/(1-4/5)) #
# " "= (1/(2/5)) + (1/(1/5)) #
# " "= 5/2 + 5 #
# " "= 15/2 #

Related questions
See all questions in Partial Sums of Infinite Series
Impact of this question
12741 views around the world
You can reuse this answer
Creative Commons License