Question

在等差数列中，试证明Sm＝p,Sp＝m,Sm＋P＝_（m＋p）

Answer 1

a(n) = a + (n-1)d.
s(n) = na + n(n-1)d/2.

p = s(m) = ma + m(m-1)d/2. p^2 = mpa + mp(m-1)d/2.
m = s(p) = pa + p(p-1)d/2. m^2 = mpa + mp(p-1)d/2.

p^2 - m^2 = [mpa + mp(m-1)d/2] - [mpa + mp(p-1)d/2] = mpd/2(m-p) = (p-m)(p+m).

m不等于p时，
mpd/2 = -(p+m).
d/2 = -(p+m)/(mp).

p = ma + m(m-1)d/2 = ma - m(m-1)(p+m)/(mp) = ma - (m-1)(p+m)/p.
a = [p + (m-1)(p+m)/p]/m.

s(m+p)= (m+p)a + (m+p)(m+p-1)d/2
= (m+p)[p+(m-1)(p+m)/p]/m + (m+p)(m+p-1)[-(p+m)/(mp)]
= [(m+p)/(mp)][p^2 + (m-1)(p+m) - (m+p-1)(m+p)]
= [(m+p)/(mp)][p^2 + (p+m)(-p)]
= [(m+p)/(mp)](-mp)
= -(m+p)