已知数列1/(1*2),1/(2*3),1/(3*4),...,1/,...,猜想Sn的表达式,并证明（归纳法）

Question

已知数列1/(1*2),1/(2*3),1/(3*4),...,1/,...,猜想Sn的表达式,并证明（归纳法）

Answer 1

猜想：Sn=n/(n+1)
证：
n=1时,S1=1/(1×2)=1/2=1/(1+1),猜想的表达式成立.
假设当n=k(k∈N,且k≥1)时,表达式成立,即Sk=k/(k+1),则当n=k+1时,
S(k+1)=1/(1×2)+1/(2×3)+...+1/[k(k+1)]+1/[(k+1)(k+2)]
=Sk+1/[(k+1)(k+2)]
=k/(k+1)+1/[(k+1)(k+2)]
=[k(k+2)+1]/[(k+1)(k+2)]
=(k²+2k+1)/[(k+1)(k+2)]
=(k+1)²/[(k+1)(k+2)]
=(k+1)/(k+2)
=(k+1)/[(k+1)+1]
表达式同样成立.
综上,得Sn的表达式为Sn=n/(n+1).