Start from the geometric series:
sum_(n=0)^oo x^n = 1/(1-x)∞∑n=0xn=11−x
for absx < 1|x|<1.
Multiply by xx:
x/(1-x) = sum_(n=0)^oo x^(n+1)x1−x=∞∑n=0xn+1
Differentiate term by term:
d/dx(x/(1-x)) = sum_(n=0)^oo d/dx( x^(n+1))ddx(x1−x)=∞∑n=0ddx(xn+1)
((1-x)+x)/(1-x)^2= sum_(n=0)^oo (n+1) x^n(1−x)+x(1−x)2=∞∑n=0(n+1)xn
1/(1-x)^2= sum_(n=0)^oo (n+1) x^n1(1−x)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(1−x)2)=∞∑n=0(n+1)ddx(xn)
-2/(1-x)^3 = sum_(n=1)^oo n(n+1) x^(n-1)−2(1−x)3=∞∑n=1n(n+1)xn−1
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(1−23)3=∞∑n=1n(n+1)(23)n−1
-2/(-1/3)^3 = 3/2 sum_(n=1)^oo n(n+1) (2/3)^n−2(−13)3=32∞∑n=1n(n+1)(23)n
2*3^3 = 3/2 sum_(n=1)^oo (n(n+1)2^n) /3^n2⋅33=32∞∑n=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^n36−−43−249−9627=∞∑n=4n(n+1)2n3n
36 - (12+ 24+ 32)/9 = sum_(n=4)^oo (n(n+1)2^n) /3^n36−12+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