若（ax+1）5的展开式中x3的系数是80，则实数a的值是______．

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问题填空题

若（ax+1）⁵的展开式中x³的系数是80，则实数a的值是______．

答案

由题意可得，T_r+1=C₅^r（ax）^5-r=a^5-rC₅^rx^5-r

令5-r=3可得r=2

∴a³C₅²=80∴a=2

故答案为：2

完形填空

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