已知{an}是等差数列,Sn为数列{an}的前n项,m,n∈N,m≠n

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已知{an}是等差数列,Sn为数列{an}的前n项,m,n∈N,m≠n
1.am=n,an=m,求am+n
2.Sn=m,Sm=n,求Sm+n

a(n) = a + d(n-1),
S(n) = na + dn(n-1)/2,
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1,
m = a + d(n-1),
n = a + d(m-1),
m - n = d(n-m),
m≠n,
d = -1.
m = a - (n-1),
a = m+n-1.
a(m+n) = a -(m+n-1) = 0.
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2,
m = na + dn(n-1)/2,
n = ma + dm(m-1)/2,
m - n = a(n-m) + d/2[n^2 - n - m^2 + m] = a(n-m) + d/2[(n-m)(n+m-1)]
= (n-m)[a + d/2(n+m-1)],
0 = a + d/2(n+m-1) + 1.
a = -1 - d/2(n+m-1)
m = na + dn(n-1)/2 = dn(n-1)/2 - n - nd/2(n+m-1) = nd/2[n-1 - n - m + 1] - n = -mnd/2 -n,
d = -2(n+m)/(mn),
a = -1 - d/2(m+n-1) = -1 +(m+n)/(mn)*(m+n-1)
S(m+n) = (m+n)a + d(m+n)(m+n-1)/2
= -m-n +(m+n)^2(m+n-1)/(mn) -2(n+m)^2(m+n-1)/(2mn)
= -m-n

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