1
an=[(n+2)/(n(n+1))]/2^(n-2)
(n +2)/(n(n+1))=(n+2)/n-(n+2)/(n+1)=2/n-1/(n+1)
an=bn+cn donde bn=(2/n)/2^(n-2)=(1/n)/2^(n-3) cn= -[1/(n+1)]/2 ^(n-2)
bn-1=(1/(n-1))/2^(n-4) cn-1=(-1/n)/2^(n- 3)
bn-2=(1/(n-2))/2^(n-5) cn-2=(-1/(n-1))/2^(n- 4)
...
b3=(1/3)/2^0 c3=(-1/4)/2
b2=( 1/2)/2^(-1) c2=(-1/3)/2^0
b1=1/2^(-2) c1=(-1/2)/2 ^(-1)
Observa que bn+cn-1=0
Sn=cn+b1=1/2^(-2)+((-1)/( n+1))/2^(n-2)
2
2×4×6×..×2n/[1×3×5×..×( 2n-1)]
=(2×4×6×..×2n)^2/[1×2×3×..×(2n-1)×2n]
=4*(n!)^2 / (2n)!