Find the value of the taylor series sum: #sum_{n=4}^∞ \frac{(n+1)(n)2^n}{3^n}# ?

#sum_{n=4}^∞ \frac{(n+1)(n)2^n}{3^n}#
(find value of taylor series sum)

1 Answer
Andrea S.
Jun 7, 2018

#sum_(n=4)^oo (n(n+1)2^n) /3^n = 256/9#

Explanation:

Start from the geometric series:

#sum_(n=0)^oo x^n = 1/(1-x)#

for #absx < 1#.

Multiply by #x#:

#x/(1-x) = sum_(n=0)^oo x^(n+1)#

Differentiate term by term:

#d/dx(x/(1-x)) = sum_(n=0)^oo d/dx( x^(n+1))#

#((1-x)+x)/(1-x)^2= sum_(n=0)^oo (n+1) x^n#

#1/(1-x)^2= sum_(n=0)^oo (n+1) x^n#

and again:

#d/dx(1/(1-x)^2)= sum_(n=0)^oo (n+1) d/dx (x^n)#

#-2/(1-x)^3 = sum_(n=1)^oo n(n+1) x^(n-1)#

Let #x=2/3#. as #0 < 2/3 < 1# this point is in the interval of convergence, so:

#-2/(1-2/3)^3 = sum_(n=1)^oo n(n+1) (2/3)^(n-1)#

#-2/(-1/3)^3 = 3/2 sum_(n=1)^oo n(n+1) (2/3)^n#

#2*3^3 = 3/2 sum_(n=1)^oo (n(n+1)2^n) /3^n#

#36 = sum_(n=1)^oo (n(n+1)2^n) /3^n#

Extract now the terms for #n=1#, #n=2# and #n=3# from the sum:

#36 = 4/3 + 24/9 + 96/27 + sum_(n=4)^oo (n(n+1)2^n) /3^n#

#36 - -4/3- 24/9 - 96/27 = sum_(n=4)^oo (n(n+1)2^n) /3^n#

#36 - (12+ 24+ 32)/9 = sum_(n=4)^oo (n(n+1)2^n) /3^n#

#256/9 = sum_(n=4)^oo (n(n+1)2^n) /3^n#

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