>>222
漸化式を(n+1)(n+2)で割って変形すると　
a[n+1]/{(n+1)(n+2)} - a[n]/{n(n+1)} = 3/(n+1) - 3/(n+2)
和を考えることで、a[n]/{n(n+1)} - a[1]/2 = 3/2 - 3/(n+1)
a[n]=n(2n-1)
T[n]=n! * (2n-1)!! = (2n)!/2^n
T[k+1]/(2k+1)! = (2k+2)/2^(k+1) =(k+1)/2^k
Σ[k=1,n] (k+1)/2^k = S と置くと
S-(1/2)S = Σ[k=1,n] (k+1)/2^k - Σ[k=1,n] (k+1)/2^(k+1)
= Σ[k=1,n] (k+1)/2^k - Σ[k=2,n+1] k/2^k
= 1 + {Σ[k=2,n] 1/2^k } - (n+1)/2^(n+1)
= 1 + 1/2 - 1/2^n - (n+1)/2^(n+1)

S = 3 - (n+3)/2^n