高中数学 数列问题 在数列{an}中,a1=1,2a（n+1）=(1+1/n)^2*an

Question

(1)求数列{an}的通项公式;
(2)令bn=a(n+1)-1/2an,求数列{bn}的前n相和Sn
(3)求数列{an}得前n项和Tn

Answer 1

解：
(1)
2a(n+1)=(1+ 1/n)²an=[(n+1)/n]²an=(n+1)²an/n²
2a(n+1)/(n+1)²=an/n²
[a(n+1)/(n+1)²]/(an/n²)=1/2，为定值
a1/1²=1/1=1，数列{an/n²}是以1为首项，1/2为公比的等比数列
an/n²=1×(1/2)^(n-1)=1/2^(n-1)
an=n²/2^(n-1)
n=1时，a1=1²/1=1，同样满足通项公式
数列{an}的通项公式为an=n²/2^(n-1)
(2)
bn=a(n+1) -(1/2)an=n²/2^(n-1) -(1/2)(n-1)²/2^(n-2)=(2n-1)/2^(n-1)
Sn=b1+b2+...+bn=1/1+3/2+5/2²+...+(2n-1)/2^(n-1)
Sn /2=1/2+3/2²+...+(2n-3)/2^(n-1)+(2n-1)/2ⁿ
Sn -Sn /2=Sn /2=1+2/2+2/2²+...+2/2^(n-1) -(2n-1)/2ⁿ
=2[1+1/2+1/2²+...+1/2^(n-1)] -(2n-1)/2ⁿ -1
=2×(1-1/2ⁿ)/(1-1/2) -(2n-1)/2ⁿ -1
=3-(2n+3)/2ⁿ
Sn=6- 2×(2n+3)/2ⁿ=6- n/2^(n-2) - 3/2^(n-1)

Answer 2

2a(n+1)=(1+1/n)^2a(n)
2^(n+1)a(n+1)=(n+1)^2[2^na(n)]/n^2
2^(n+1)a(n+1)/(n+1)^2=2^na(n)/n^2=...=2a(1)/1=2
a(n)=n^2/2^(n-1)
b(n)=a(n+1)=(n+1)^2/2^n=n(n+1)/2^n + (n+1)/2^n=c(n)+d(n)
c(n)=n(n+1)/2^n, d(n)=(n+1)/2^n
D(n)=d(1)+d(2)+...+d(n)=2/2 + 3/2^2 + 4/2^3 + ... + (n-1+1)/2^(n-1) + (n+1)/2^n
2D(n)=2 + 3/2 + 4/2^2 +...+ (n-1+1)/2^(n-2) + (n+1)/2^(n-1)
D(n)=2D(n)-D(n)=2+1/2+1/2^2+...+1/2^(n-1) - (n+1)/2^n
=1-(n+1)/2^n + 2[1-1/2^n]
=3-(n+3)/2^n
C(n)=c(1)+c(2)+...+c(n)=1*2/2 + 2*3/2^2 + 3*4/2^3 + ...+(n-1)n/2^(n-1) + n(n+1)/2^n
2C(n)=1*2 + 2*3/2 + 3*4/2^2 + ... + (n-1)n/2^(n-2) + n(n+1)/2^(n-1)
C(n)=2C(n)-C(n)=1*2 + 2(3-1)/2 + 3(4-2)/2^2 + ... + n(n+1-n+1)/2^(n-1) - n(n+1)/2^n
=2+ 2[(1+1)/2 + (2+1)/2^2 + ... + (n-1+1)/2^(n-1)] - n(n+1)/2^n
=2+2[D(n)-(n+1)/2^n] - n(n+1)/2^n
S(n)=C(n)+D(n)=2+2[D(n)-(n+1)/2^n] - n(n+1)/2^n + D(n)
=2-n(n+1)/2^n - 2(n+1)/2^n + 3D(n)
=2-n(n+1)/2^n -2(n+1)/2^n + 3[3-(n+3)/2^n]
=11-(n^2+6n+11)/2^n

Answer 3

问数学老师

Answer 4

什么没什么是什么什么什么什么