Find the value of the taylor series sum: sum_{n=4}^∞ \frac{(n+1)(n)2^n}{3^n}n=4(n+1)(n)2n3n ?

sum_{n=4}^∞ \frac{(n+1)(n)2^n}{3^n}n=4(n+1)(n)2n3n
(find value of taylor series sum)

1 Answer
Jun 7, 2018

sum_(n=4)^oo (n(n+1)2^n) /3^n = 256/9n=4n(n+1)2n3n=2569

Explanation:

Start from the geometric series:

sum_(n=0)^oo x^n = 1/(1-x)n=0xn=11x

for absx < 1|x|<1.

Multiply by xx:

x/(1-x) = sum_(n=0)^oo x^(n+1)x1x=n=0xn+1

Differentiate term by term:

d/dx(x/(1-x)) = sum_(n=0)^oo d/dx( x^(n+1))ddx(x1x)=n=0ddx(xn+1)

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

1/(1-x)^2= sum_(n=0)^oo (n+1) x^n1(1x)2=n=0(n+1)xn

and again:

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

-2/(1-x)^3 = sum_(n=1)^oo n(n+1) x^(n-1)2(1x)3=n=1n(n+1)xn1

Let x=2/3x=23. as 0 < 2/3 < 10<23<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(123)3=n=1n(n+1)(23)n1

-2/(-1/3)^3 = 3/2 sum_(n=1)^oo n(n+1) (2/3)^n2(13)3=32n=1n(n+1)(23)n

2*3^3 = 3/2 sum_(n=1)^oo (n(n+1)2^n) /3^n233=32n=1n(n+1)2n3n

36 = sum_(n=1)^oo (n(n+1)2^n) /3^n36=n=1n(n+1)2n3n

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

36 = 4/3 + 24/9 + 96/27 + sum_(n=4)^oo (n(n+1)2^n) /3^n36=43+249+9627+n=4n(n+1)2n3n

36 - -4/3- 24/9 - 96/27 = sum_(n=4)^oo (n(n+1)2^n) /3^n36432499627=n=4n(n+1)2n3n

36 - (12+ 24+ 32)/9 = sum_(n=4)^oo (n(n+1)2^n) /3^n3612+24+329=n=4n(n+1)2n3n

256/9 = sum_(n=4)^oo (n(n+1)2^n) /3^n2569=n=4n(n+1)2n3n