已知数列｛an｝首项a1=1，递推公式an=an-1+2，求通项公式an 已知数列{an}中,a1=1,以后各项由递推公式an+1=a...
\u6839\u636e\u6570\u5217{an}\u7684\u9996\u9879a1=1\uff0c\u548c\u9012\u63a8\u5173\u7cfban=2an-1+1\uff0c\u63a2\u6c42\u5176\u901a\u9879\u516c\u5f0f\u4e3a______\u2235an=2an-1+1\uff0c\u2234an+1=2an-1+1+1=2\uff08an-1+1\uff09\uff0c\u5219\u6570\u5217{an+1}\u662f\u516c\u6bd4q=2\u7684\u7b49\u6bd4\u6570\u5217\uff0c\u9996\u9879\u4e3aa1+1=1+1=2\uff0c\u5219an+1=2\u00d72n-1=2n\uff0c\u5219an=2n-1\uff0c\u6545\u6570\u5217\u7684\u901a\u9879\u516c\u5f0f\u4e3aan=2n-1\uff0c\u6545\u7b54\u6848\u4e3a\uff1aan=2n-1
\u7531A\uff08n+1\uff09=An+2\u5f97
A\uff08n+1\uff09-An=2
\u663e\u7136\u5217{An}\u662f\u4ee5A1=1\u4e3a\u9996\u9879\uff0cd=2\u4e3a\u516c\u5dee\u7684\u7b49\u5dee\u6570\u5217\uff0c\u6240\u4ee5\u6570\u5217{An}\u7684\u901a\u9879\u516c\u5f0f\u4e3a
An=1+2\uff08n-1\uff09=2n-1
依据等差数列的通项公式可知
an=a1+(n-1)d
=1+(n-1)×2=2n-1
所以通项公式为an=2n-1
绛旓細an=a1+(n-1)d =1+(n-1)脳2=2n-1 鎵浠ラ氶」鍏紡涓篴n=2n-1
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