已知数列an中，a1=1，an+1=an+n，求an
\u8bbe\u6570\u5217{an}\u4e2d,a1=2,a(n+1)=an+n+1,\u6c42an\u6709\u9898\u76ee\u53ef\u77e5,a(n+1)-an=n+1\uff0c\u6240\u4ee5
a2-a1=2,
a3-a2=3,
a4-a3=4
an-a(n-1)=n
\u628a\u4ee5\u4e0a\u5f0f\u5b50\u5168\u52a0\u8d77\u6765\u7684\u5230an-a1=2+3+4+~~~n,\u6240\u4ee5an=2+2+3+4+~~~n=(n+1)n/2+1\uff0c\u6b64\u4e2d\u7684n\u662f\u5927\u4e8e\u4e00\u7684\uff0c\u5f53n\u7b49\u4e8e\u4e00\u65f6\u5bf9\u4e8ea1\u4e5f\u662f\u6210\u7acb\u7684\u6240\u4ee5an=(n+1)n/2+1
\u4f60\u63d0\u4f9b\u7684\u7b54\u6848\u5e94\u8be5\u662f\u9519\u8bef\u7684\u4f60\u53ef\u4ee5\u5427n=1\u5e26\u5165\u4f60\u7684\u7b54\u6848\u4e2d\u53d1\u73b0a1=5\u4e0e\u9898\u76ee\u7ed9\u4f60\u7684\u6761\u4ef6\u662f\u4e0d\u7b26\u5408\u7684\u3002
a(n\uff0b1)\uff0dan=(1/2)[1/n\uff0d1/(n\uff0b2)]\uff0c\u5219\uff1a
a2\uff0da1=(1/2)[1/1\uff0d1/3]
a3\uff0da2=(1/2)[1/2\uff0d1/4]
a4\uff0da2=(1/2)[1/3\uff0d1/5]
\u2026\u2026
an\uff0da(n\uff0d1)=(1/2)[1/(n\uff0d1)\uff0d1/(n\uff0b1)]
\u5168\u90e8\u76f8\u52a0\uff0c\u5f97\uff1a
an\uff0da1=(1/2)[1\uff0b1/2\uff0d1/n\uff0d1/(n\uff0b1)]
\u6240\u4ee5\uff0can=2\uff0b(1/2)[3/2\uff0d(2n\uff0b1)/(n²\uff0bn)]
一次前后累加，an-a1=(n-1)*n/2,所以an=(n-1)*n/2+1。
A2=A1+1
A3=A2+2
A4=A3+3
..............
An=A(n-1)+(N-1)
左式上下相加=右式上下相加
An=A1+[1+2+3+...+(N-1)]
An=1+[N(N-1)]/2
an=1+2+3+....+n-1=n*(n-1)/2
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