a(1)=5/6,
a(n+1) = a(n)/3 + (1/2)^(n+1),
2^(n+1)a(n+1) = 2^na(n)*(2/3) + 1,
记b(n) = 2^na(n),则
b(n+1) = 2^(n+1)a(n+1) = 2^na(n)*(2/3) + 1 = 2b(n)/3 + 1,
b(n+1) - 3 = 2b(n)/3 - 2 = (2/3)[b(n) - 3],
{b(n) - 3}是首项为b(1)-3 = 2a(1)-3 = 5/3-3 = -4/3, 公比为（2/3）的等比数列.
b(n) - 3 = (-4/3)(2/3)^(n-1) = -2(2/3)^n = 2^na(n) - 3,
2^na(n) = 3 - 2(2/3)^n,
a(n) = 3/2^n - 2/3^n
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a(1)=1/2,
a(n+1) = a(n) + 1/(n^2 + n) = a(n) + 1/[n(n+1)] = a(n) + 1/n - 1/(n+1),
a(n+1) + 1/(n+1) = a(n) + 1/n,
{a(n)+1/n}是首项为a(1)+1 = 3/2,的常数数列.
a(n) + 1/n = 3/2,
a(n) = 3/2 - 1/n.
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试题上的解题方法详解如下：
a(n+1) = pa(n) + rq^n,
先消去q^n项.
当q不为0时,
a(n+1)/q^n = (p/q)*a(n)/q^(n-1) + r,
记b(n) = a(n)/q^(n-1),则有,
b(n+1) = (p/q)b(n) + r,
若p/q = 1,则b(n+1) = b(n) + r, {b(n)}是首项为b(1)=a(1),公差为r的等差数列.
b(n) = a(1) + (n-1)r = a(n)/q^(n-1),
a(n) = [a(1) - r + nr]q^(n-1).
p/q不为1时,
b(n+1) = (p/q)b(n) + r,
再消去r...
b(n+1) + r/[p/q - 1] = (p/q)b(n) + r + r/[p/q - 1] = (p/q)b(n) + r(p/q)/[p/q - 1]
= (p/q){ b(n) + r/[p/q - 1] },
{b(n) + r/[p/q-1]}是首项为b(1) + r/[p/q-1] = a(1) + r/(p/q-1),公比为(p/q)的等比数列.
b(n) + r/[p/q-1] = [a(1) + r/(p/q-1)](p/q)^(n-1),
b(n) = [a(1) + r/(p/q-1)](p/q)^(n-1) - r/(p/q-1) = a(n)/q^(n-1),
a(n) = [a(1) + r/(p/q-1)]p^(n-1) - rq^(n-1)/(p/q-1)
= [a(1) + rq/(p-q)]p^(n-1) - rq^n/(p-q)
