Question

已知各项大于零的数列｛ak}的前k项和为Sk,且∑(上面是n,下面是k=1）ak^3(k为下标）=Sn^2,求数列通项

Answer 1

经计算可得a(1)=1,由已知条件有S(n+1)^2 = ∑(k=1至n+1)a(k)^3 = S(n)^2 + a(n+1)^3,所以

a(n+1)^3 = (S(n+1) - S(n)) * (S(n+1) + S(n)) = a(n+1) * (a(n+1) + 2S(n)),

进而可得a(n+1)^2 - a(n+1) -2S(n) =0.(#)

接下来用数学归纳法证明a(n)=n.当n=1时显然有a(1)=1,假设n≤k时结论成立,则

S(k) = a(1) + a(2) + ...+ a(k) = 1 + 2 + ...+ k = k(k+1)/2,

于是由(#)式可得a(k+1)^2 - a(k+1) - k(k+1) = 0,因为a(k+1)>0,所以a(k+1) = k+1.

所以n=k+1时a(n)=n也成立.所以数列的通项就是a(n) = n.