2pSn=an^2+pan 数列an各项都为正数p大于0为常数,记1/SN的前n项和为Hn,

2ps(n) = [a(n)]^2 + pa(n), p>0. a(n)>0.
2pa(1) = 2ps(1) = [a(1)]^2 + pa(1),
0 = [a(1)]^2 -pa(1) = a(1)[a(1)-p],
a(1) = p.
2ps(n+1) = [a(n+1)]^2 + pa(n+1),
2pa(n+1) = 2ps(n+1)-2ps(n) = [a(n+1)]^2 + pa(n+1) - [a(n)]^2 - pa(n),
0 = [a(n+1)]^2 - [a(n)]^2 - pa(n+1) - pa(n) = [a(n+1)+a(n)][a(n+1)-a(n) - p],
0 = a(n+1) - a(n) - p,
a(n+1) = a(n) + p,
{a(n)}是首项为a(1)=p,公差为p的等差数列.
a(n) = p + (n-1)p = np.
2ps(n) = [a(n)]^2 + pa(n) = p^2n^2 + p*np = p[pn^2 + np],
s(n) = pn(n+1)/2.
1/s(n) = (2/p)/[n(n+1)] = (2/p)[1/n - 1/(n+1)],
h(n) = 1/s(1) + 1/s(2) + ... + 1/s(n-1) + 1/s(n) = (2/p)[ 1-1/2 + 1/2-1/3 + ... + 1/(n-1) - 1/n + 1/n - 1/(n+1)] = (2/p)[ 1 - 1/(n+1)] = 2n/[p(n+1)]

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