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

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 intege