Find the value of the taylor series sum: n=4(n+1)(n)2n3n ?

n=4(n+1)(n)2n3n
(find value of taylor series sum)

1 Answer
Jun 7, 2018

n=4n(n+1)2n3n=2569

Explanation:

Start from the geometric series:

n=0xn=11x

for |x|<1.

Multiply by x:

x1x=n=0xn+1

Differentiate term by term:

ddx(x1x)=n=0ddx(xn+1)

(1x)+x(1x)2=n=0(n+1)xn

1(1x)2=n=0(n+1)xn

and again:

ddx(1(1x)2)=n=0(n+1)ddx(xn)

2(1x)3=n=1n(n+1)xn1

Let x=23. as 0<23<1 this point is in the interval of convergence, so:

2(123)3=n=1n(n+1)(23)n1

2(13)3=32n=1n(n+1)(23)n

233=32n=1n(n+1)2n3n

36=n=1n(n+1)2n3n

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

36=43+249+9627+n=4n(n+1)2n3n

36432499627=n=4n(n+1)2n3n

3612+24+329=n=4n(n+1)2n3n

2569=n=4n(n+1)2n3n