級數和的問題

Question

求Sn=1/1*3+1/3*5+1/5*7+...+1/(2n-1)*(2n+1)
謝謝~

請副上算式，謝謝~

Answer 1

2Sn=2/1*3+2/3*5+2/5*7+...+2/(2n-1)*(2n+1)
=1/1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-1)-1/(2n+1)
中間(-1/3+1/3-1/5+1/5-1/7+...+1/(2n-1)...)可以消去
只剩下1-1/(2n+1)兩項
2Sn= 1-1/(2n+1) = 2n/(2n+1)
所以 Sn= n/(2n+1)

2005-09-20 23:47:13 補充：
1/1-1/3=2/1*3, 1/3-1/5=2/3*5以此類推囉

Answer 2

Sn=1/1*3+1/3*5+1/5*7+...+1/(2n-1)*(2n+1)
=(1/2)[(1/1)-(1/3)]+(1/2)[(1/3)-(1/5)]+(1/2)[(1/5)-(1/7)]+...+(1/2)[1/(2n-1)-1/(2n+1)]
=(1/2)[(1/1)-(1/3)+(1/3)-(1/5)+(1/5)-(1/7)+...+1/(2n-1)-1/(2n+1)]
=(1/2)[(1/1)-1/(2n+1)]
=(1/2)[2n/(2n+1)]
=n/(2n+1)

Answer 3

為什麼=1/1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-1)-1/(2n+1)

Answer 4

Sn=n/(2n+1) 當n=1時，1/3=1/(2*1+1)成立假設當n=k時成立，即1/(1*3_+1/(3*5)+...+1/[(2k-1)*(2k+1)]=k/(2k+1)當n=k+1時， 1/(1*3)+1/(3*5)+...+1/[(2k-1)*(2k+1)]+1/[(2k+1)*(2k+3)]=k/(2k+1)+1/[(2k+1)*(2k+3)]=[k(2k+3)+1]/[(2k+1)*(2k+3)]=(2k2+3k+1)/[(2k+1)*(2k+3)]=[(2k+1)*(k+1)]/[(2k+1)*(2k+3)]=(k+1)/[2(k+1)+1]也成立得證