Start from the geometric series:
sum_(n=0)^oo q^n = 1/(1-q)∞∑n=0qn=11−q 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)^2∞∑n=1(−1)nnxn−1=−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)^2∞∑n=1(−1)nn(34)n−1=−1(1+34)2
sum_(n=1)^oo (-1)^n (n3^(n-1))/2^(2n-2) = -16/49∞∑n=1(−1)nn3n−122n−2=−1649
sum_(n=1)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = -16/49∞∑n=1(−1)n(2n)3n−122n−1=−1649
Extract the term for n=1n=1:
-1+sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = -16/49−1+∞∑n=2(−1)n(2n)3n−122n−1=−1649
sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = -16/49+1∞∑n=2(−1)n(2n)3n−122n−1=−1649+1
sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = 33/49∞∑n=2(−1)n(2n)3n−122n−1=3349