数列an满足a1=2012,a n+1 +a n +n²=0 求a11
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数列an满足a1=2012,a n+1 +a n +n²=0 求a11
这怎么做啊?求解,急谢了
这怎么做啊?求解,急谢了
/>a(n+1)+an +n²=0,则
a(n+2)+a(n+1)+(n+1)²=0
减去上式得
a(n+2)=an+n²-(n+1)²=an-2n-1
于是
a11=a9-2*9-1
=a7-2*7-1-2*9-1
=a5-2*5-1-2*7-1-2*9-1
=a3-2*3-1-2*5-1-2*7-1-2*9-1
=a1-2*1-1-2*3-1-2*5-1-2*7-1-2*9-1
=2012-2(1+3+5+7+9)-5
=1957
a(n+2)+a(n+1)+(n+1)²=0
减去上式得
a(n+2)=an+n²-(n+1)²=an-2n-1
于是
a11=a9-2*9-1
=a7-2*7-1-2*9-1
=a5-2*5-1-2*7-1-2*9-1
=a3-2*3-1-2*5-1-2*7-1-2*9-1
=a1-2*1-1-2*3-1-2*5-1-2*7-1-2*9-1
=2012-2(1+3+5+7+9)-5
=1957
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