(a+x)(1+√x)^5的展开式中x2项系数为15则展开式的所有项系数的和 明天就要高考 急 (a+x) 这个怎么算啊 (a+x)(1+√x)^5?
2019-12-04
(a+x)(1+√x)^5的展开式中x2项系数为15则展开式的所有项系数的和 明天就要高考 急
(a+x) 这个怎么算啊 (a+x)(1+√x)^5?
优质解答
(a+x)(1+√x)^5
coef. of x^2
a(5C4) + 5C2 = 15
5a+10=15
a=1
ie
(1+x)(1+√x)^5
=(1+x)(1+5√x+10x+10x√x+5x^2+x^2√x)
= (1+5√x+10x+10x√x+5x^2+x^2√x) + (x+5x√x+10x^2+10x^2√x+5x^3+x^3√x)
=1+5√x+11x+15x√x+15x^2+11x^2√x+5x^3+x^3√x
所有项系数的和
=1+5+11+15+15+11+5+1
=64
(a+x)(1+√x)^5
coef. of x^2
a(5C4) + 5C2 = 15
5a+10=15
a=1
ie
(1+x)(1+√x)^5
=(1+x)(1+5√x+10x+10x√x+5x^2+x^2√x)
= (1+5√x+10x+10x√x+5x^2+x^2√x) + (x+5x√x+10x^2+10x^2√x+5x^3+x^3√x)
=1+5√x+11x+15x√x+15x^2+11x^2√x+5x^3+x^3√x
所有项系数的和
=1+5+11+15+15+11+5+1
=64