Question

设数列(an)是首项为a1(a＞0),公差为2的等差数列,其前n项和为Sn,且√s1,√s2,√s3

Answer 1

令S1=a1=t

S2=a1+a2=2a1+2=2t+2

S3=a1+a2+a3=3a1+6=3t+6

2√S2=√S1+√S3,

2√(2t+2)=√t+√(3t+6),

4(2t+2)=t+3t+6+2√[t(3t+6)]

8t+8=4t+6+2√(3t²+6t)

4t+2=2√(3t²+6t)

16t²+16t+4=4(3t²+6t)

16t²+16t+4=12t²+24t

4t²-8t+4=0

t²-2t+1=0

(t-1)²=0

t=1

即a1=1

an=a1+(n-1)d=1+2(n-1)=2n-1

∴an=2n-1

bn=(an)/(2^n)=(2n-1)/(2^n)

Tn=b1+b2+b3+...+bn

2Tn=2b1+2b2+2b3+...+2bn

2Tn-Tn

=(2b1+2b2+2b3+...+2bn)-(b1+b2+b3+...+bn)

=2b1+(2b2-b1)+(2b3-b2)+...+(2bn-b[n-1])-bn

=1+(3/2-1/2)+(5/4-3/4)+...+[(2n-1)/2^(n-1)-(2n-3)/2^(n-1)] - (2n-1)/2^n

=1+1+[1/2+1/4+...+1/2^(n-2)] - (2n-1)/2^n

=2+ [1-1/2^(n-2)] - (2n-1)/2^n

=3- (2n+3)/2^n

∴Tn=3- (2n+3)/2^n