Question

在等差数列中，试证明Sm＝p,Sp＝m,Sm＋P＝_（m＋p）

Answer 1

简单计算一下即可，答案如图所示

Answer 2

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)