若an是等差数列,求证a1(2^)-a2(2^)+a3(2^)-a4(2^)+a2n-1(2^)-a2n(2^)=n/2n-1[a1(2^)-a2n(2^)]

若an是等差数列,求证a1(2^)-a2(2^)+a3(2^)-a4(2^)+a2n-1(2^)-a2n(2^)=n/2n-1[a1(2^)-a2n(2^)]
若an是等差数列,求证a1(2^)-a2(2^)+a3(2^)-a4(2^)+a2n-1(2^)-a2n(2^)=n/2n-1[a1(2^)-a2n(2^)]

若an是等差数列,求证a1(2^)-a2(2^)+a3(2^)-a4(2^)+a2n-1(2^)-a2n(2^)=n/2n-1[a1(2^)-a2n(2^)]
等式右边=-[a2(2^)-a1(2^)+a4(2^)-a3(2^）+······+a2n(2^)-a2n-1(2^)]
=-[(a2-a1) (a2+a1)+(a4-a3) (a4+a3)+······+(a2n-a2n-1) (a2n+a2n-1)]
=-d(a1+a2+a3+a4+······+a2n-1+a2n)
=-d [2n(an+a2n)/2]
=-[(an-a1)/2n-1] [2n(a1+a2n)/2]
=[(a1-an)/2n-1][n(a1+a2n)]
=n/2n-1[a1(2^)-a2n(2^)]
此题的关键是d=am-an/m-n ,Sn=n（a1+an)/2 以及平方差公式的运用.