a(n)=aq^(n-1),a>0,q>0.
a+aq=a(1)+a(2)=2[1/a(1)+1/a(2)]=2[1/a+1/(aq)]=2(q+1)/(aq),
a=2/(aq),q=2/a^2,
a(n)=a*(2/a^2)^(n-1)=2^(n-1)a^(3-2n),
4a^(-3)+8a^(-5)+16a^(-7)=64[1/a(3)+1/a(4)+1/a(5)]=64[1/4a^3+1/8a^5+1/16a^7]=16a^3+8a^5+4a^7,
a^(-3)+2a^(-5)+4a^(-7)=4a^3+2a^5+a^7,
a^4+2a^2+4=4a^(10)+2a^(12)+a^(14),
0=a^(14)+2a^(12)+4a^(10)-a^4-2a^2-4=a^(13)[a-1]+a^(12)[a-1]+3a^(11)[a-1]+3a^(10)[a-1]+7a^9[a-1]+7a^8[a-1]+7a^7[a-1]+7a^6[a-1]+7a^5[a-1]+7a^4[a-1]+6a^3[a-1]+6a^2[a-1]+4a[a-1]+4[a-1]
=[a^(13)+a^(12)+3a^(11)+3a^(10)+7a^9+7a^8+7a^7+7a^6+7a^5+7a^4+6a^3+6a^2+4a+4][a-1],
因a>0,所以
a^(13)+a^(12)+3a^(11)+3a^(10)+7a^9+7a^8+7a^7+7a^6+7a^5+7a^4+6a^3+6a^2+4a+4>0,
所以,a=1.
a(n)=2^(n-1),n=1,2,..
b(n)=[a(n)+1/a(n)]^2=[2^(n-1)+2^(1-n)]^2=4^(n-1)+4^(1-n)+2,
T(n)=b(1)+b(2)+...+b(n)=[1+4+...+4^(n-1)]+[1+1/4+...+1/4^(n-1)]+2n
=[4^n-1]/(4-1)+[1-1/4^n]/[1-1/4]+2n
=(1/3)[4^n-1]+(4/3)[1-1/4^n]+2n
=(1/3)4^n - (1/3)4^(1-n) +2n + 1,
如果要用b(n)=T(n)-T(n-1)來(lái)求T(n)的話(huà),
4^(n-1)+4^(1-n)+2=b(n)=T(n)-T(n-1),
4^(n-2)+4^(2-n)+2=T(n-1)-T(n-2),
...
4+4^(-1)+2=T(2)-T(1),
[4^(n-1)+4^(n-2)+...+4] + [4^(1-n)+4^(2-n)+...+4^(-1)] + 2(n-1) = T(n)-T(1),
4[4^(n-1)-1]/(4-1) + (1/4)[1-4^(1-n)]/(1-1/4) + 2(n-1) = T(n)-b(1)=T(n)-4,
T(n)=(1/3)[4^n-4] + (1/3)[1-4^(1-n)] + 2(n-1) + 4
=(1/3)4^n - (1/3)4^(1-n) + 2n + 1,
答案也是一樣.
分組求和 == 前n項(xiàng)和