在等差数列an中,各项均不为0,求证:1/a1×a2+1/a2×a3+...+1/an×an+1=n/a1×an+1
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证：
数列公差d=0时,即数列各项均相等时,
1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]=n/a1²
n/[a1a(n+1)]=n/a1²
1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]=n/[a1a(n+1)],等式成立.
公差d≠0时,
1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]
=(1/d)[1/a1 -1/a2 +1/a2 -1/a3+...+1/an -1/a(n+1)]
=(1/d)[1/a1 -1/a(n+1)]
n/[a1a(n+1)]=n/[a1(a1+nd)]=n×[1/(nd)][1/a1 - 1/a(n+1)]=(1/d)[1/a1 -1/a(n+1)]
1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]=n/[a1a(n+1)],等式成立.
综上,得1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]=n/[a1a(n+1)],等式成立.
提示：用到一个基本的变换：1/[a(a+d)]=(1/d)[1/a -1/(a+d)]
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