已只（X+1）的ｎ次展开式中第2第5项系数相等、求第四项

Question

已只（X+1）的ｎ次展开式中第2第5项系数相等、求第四项

Answer 1

(x + 1) n 的展开式的通项公式为：T r+1 = C n r *x (n – r) *1 r = C n r *x (n – r) ,所以第二项的系数为C n 1 ,第五项的系数为C n 4 ,所以C n 1 = C n 4 => n = n(n – 1)(n – 2)(n – 3)/(1*2*3*4) => (n – 1)(n – 2)(n – 3) = 24 => n 3 –6n 2 + 11n – 30 = 0 => (n – 5)(n 2 –n + 6) = 0 => n – 5 = 0 => n = 5 => 第四项为T 4 = C n 3 *x (n – 3) = C 5 3 *x (5 – 3) = 10x 2 .