已知数列8*1/1^2*3^2,8*2/3^2*5^2,……8*n/(2n-1)^2(2n+1)^2,…,Sn为该数列的前n项和,计算得S1=8/9,S2=24/25,S3=48/49,S4=80/81.观察上述结果,推测出Sn(n∈N*),并用数学归纳法加以证明

Question

已知数列8*1/1^2*3^2,8*2/3^2*5^2,……8*n/(2n-1)^2(2n+1)^2,…,Sn为该数列的前n项和,计算得S1=8/9,S2=24/25,S3=48/49,S4=80/81.观察上述结果,推测出Sn(n∈N*),并用数学归纳法加以证明

Answer 1

an = 8n/[(2n-1)(2n+1)]^2
= 1/(2n-1)^2 - 1/(2n+1)^2
Sn = a1+a2+...+an
= 1 - 1/(2n+1)^2
By MI
Sn = 1 - 1/(2n+1)^2
n=1
S1 = 8/9 = 1- 1/(2+1)^2
p(1) is true
Assume p(k) is true
ie
Sk =1 - 1/(2k+1)^2
for n=k+1
S(k+1) = 1 - 1/(2k+1)^2 + 8(k+1)/[(2k+1)(2k+3)]^2
= 1 - [(2k+3)^2 -8(k+1)]/[(2k+1)(2k+3)]^2
= 1 - (4k^2+4k+1)/[(2k+1)(2k+3)]^2
= 1- 1/(2k+3)^2
p(k+1) is true
By principle of MI,it is true for all +ve integer