Evaluate the taylor series sum: sum_{n=2}^{∞} {(-1)^n 3^{n-1}(2n)}/(2^{2n-1})n=2(1)n3n1(2n)22n1 ?

sum_{n=2}^{∞} {(-1)^n 3^{n-1}(2n)}/(2^{2n-1})n=2(1)n3n1(2n)22n1 use taylor series to evaluate this sum.

1 Answer
Jun 22, 2018

sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = 33/49n=2(1)n(2n)3n122n1=3349

Explanation:

Start from the geometric series:

sum_(n=0)^oo q^n = 1/(1-q)n=0qn=11q for abs q < 1|q|<1

Let q=-xq=x to have:

sum_(n=0)^oo (-1)^nx^n = 1/(1+x)n=0(1)nxn=11+x for abs x < 1|x|<1

Differentiate term by term:

sum_(n=1)^oo (-1)^n nx^(n-1) = -1/(1+x)^2n=1(1)nnxn1=1(1+x)2 for abs x < 1|x|<1

Let x=3/4x=34:

sum_(n=1)^oo (-1)^n n(3/4)^(n-1) = -1/(1+3/4)^2n=1(1)nn(34)n1=1(1+34)2

sum_(n=1)^oo (-1)^n (n3^(n-1))/2^(2n-2) = -16/49n=1(1)nn3n122n2=1649

sum_(n=1)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = -16/49n=1(1)n(2n)3n122n1=1649

Extract the term for n=1n=1:

-1+sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = -16/491+n=2(1)n(2n)3n122n1=1649

sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = -16/49+1n=2(1)n(2n)3n122n1=1649+1

sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = 33/49n=2(1)n(2n)3n122n1=3349