若lim[2n+(an^2+2n+1)/(bn+1)=1,则a+b
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若lim[2n+(an^2+2n+1)/(bn+1)=1,则a+b若lim[2n+(an^2+2n+1)/(bn+1)=1,则a+b若lim[2n+(an^2+2n+1)/(bn+1)=1,则a+bl
若lim[2n+(an^2+2n+1)/(bn+1)=1,则a+b
若lim[2n+(an^2+2n+1)/(bn+1)=1,则a+b
若lim[2n+(an^2+2n+1)/(bn+1)=1,则a+b
lim(n->inf) [2n + (an² + 2n + 1) / (bn + 1)] = 1
lim (2bn² + an² + 4n + 1) / (bn + 1) = 1
lim [(2b + a)n² + 4n + 1] / (bn + 1) = 1
2b + a = 0,∵极限趋向常数,分子和分母的最高次方相同,都是n的一次方
lim (4n + 1) / (bn + 1) = 1
lim (4 + 1/n) / (b + 1/n) = 1
(4 + 0) / (b + 0) = 1
b = 4
a = -2b = -8
∴a + b = -8 + 4 = -4
表述好像有点不清楚
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