我要问的是第4题!2n +(an^2-2n+1)/(bn+2)=[2n(bn+2)+(an^2-2n+1)]/(bn+2)=[(a+2b)n^2+2n+1]/(bn+2)分子包含n^2,要极限有界,则a+2b=0 a=-2blim[(2n+1)/(bn+2)]=1lim[(2+1/n)/(b +2/n)]=1n->+∞,lim[(2+1/n)/(b+2/n)]=2/b=1b=2a=-2b=-4实数对(
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我要问的是第4题!2n +(an^2-2n+1)/(bn+2)=[2n(bn+2)+(an^2-2n+1)]/(bn+2)=[(a+2b)n^2+2n+1]/(bn+2)分子包含n^2,要极限有界,则a+2b=0 a=-2blim[(2n+1)/(bn+2)]=1lim[(2+1/n)/(b +2/n)]=1n->+∞,lim[(2+1/n)/(b+2/n)]=2/b=1b=2a=-2b=-4实数对(
我要问的是第4题!
2n +(an^2-2n+1)/(bn+2)
=[2n(bn+2)+(an^2-2n+1)]/(bn+2)
=[(a+2b)n^2+2n+1]/(bn+2)
分子包含n^2,要极限有界,则a+2b=0 a=-2b
lim[(2n+1)/(bn+2)]=1
lim[(2+1/n)/(b +2/n)]=1
n->+∞,lim[(2+1/n)/(b+2/n)]=2/b=1
b=2a=-2b=-4实数对(a,b)为(-4,2).
我要问的是：怎么由=[2n(bn+2)+(an^2-2n+1)]/(bn+2)得到这个=[(a+2b)n^2+2n+1]/(bn+2)的!
还有不懂：
是2n(bn+2)=为什么会得到2bn^2+4n的！
我要问的是第4题!2n +(an^2-2n+1)/(bn+2)=[2n(bn+2)+(an^2-2n+1)]/(bn+2)=[(a+2b)n^2+2n+1]/(bn+2)分子包含n^2,要极限有界,则a+2b=0 a=-2blim[(2n+1)/(bn+2)]=1lim[(2+1/n)/(b +2/n)]=1n->+∞,lim[(2+1/n)/(b+2/n)]=2/b=1b=2a=-2b=-4实数对(
2n(bn+2)=2bn^2+4n
2n(bn+2)+an^2-2n+1=2bn^2+4n+an^2-2n+1=(a+2b)n^2+2n+1
实在是没想通这有什么不懂的。。
分子展开之后再把同项合并不就得到了，分母又没有变化。。
由匚2n(bn+2)十(an^2一2n+1]=2nbn十4n十an∧2一2n十1=(2n十bn)n^2十1
只看分子=2n(bn+2)+(an^2-2n+1)
=2bn^2+4n+an^2-2n+1
=(a+2b)n^2+2n+1
你问"是2n(bn+2)=为什么会得到2bn^2+4n的！"?(我怀疑你是不是在开玩笑..)
不就是把2n乘以括号里的每一项吗?
2n(bn+2)=2n*bn+2n*2=2bn^2+4n呵呵，看错了，以为是bn的通项，是啊！闹笑话了！<...
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只看分子=2n(bn+2)+(an^2-2n+1)
=2bn^2+4n+an^2-2n+1
=(a+2b)n^2+2n+1
你问"是2n(bn+2)=为什么会得到2bn^2+4n的！"?(我怀疑你是不是在开玩笑..)
不就是把2n乘以括号里的每一项吗?
2n(bn+2)=2n*bn+2n*2=2bn^2+4n
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