Question

高中数学数列题。第二问。求大神答疑解惑(๑°3°๑)

Answer 1

约定:[ ]内是下标
(1)由已知设a[1]=a,公差d(d≠0)
得 a+(a+d)(a+2d)=21 即a+d=7 (1)
(a+5d)²=a(a+20d) (2)
由(1)(2)解得 a=5,d=2
所以 a[n]=2n+3
(2)设c[n]=1/b[n]
则c[1]=3
且n≥2时 c[n]-c[n-1]=a[n-1]=2n+1
即c[n]-c[n-1]=2n+1
c[n]=(c[n]-c[n-1])+(c[n-1]-c[n-2])+...+(c[2]-c[1])+c[1]
=(2n+1)+(2(n-1)+1)+...+(2·2+1)+3
=n(3+(2n+1))/2
=n²+2n
b[n]=1/(n²+2n)=(1/2)((1/n)-(1/(n+2)))
T[n]=(1/2)((1-(1/3))+((1/2)-(1/4)+...+((1/(n-1))-(1/(n+1))+((1/n)-(1/(n+2)))
=(1/2)(1+(1/2)-(1/(n+1))-(1/(n+2)))
=(3/4)-(2n+3)/(2n²+6n+4)
所以 T[n]=(3/4)-(2n+3)/(2n²+6n+4)

希望能帮到你！