Question

用数学归纳法证明 1/1*2+1/3*4+…+1/(2n-1)*2n=1/(n+1)+1/(n+2)+…+1/(n+n)

Answer 1

① n=1时,左=1/1*2=1/2 ,右=1/(1+1)=1/2 ∴左=右 等式成立

②

假设 n=k 时等式成立,即有 1/1*2+1/3*4+…+1/(2k-1)*2k=1/(k+1)+1/(k+2)+…+1/(k+k)

n=k+1 时 左= 1/1*2+1/3*4+…+1/(2k-1)*2k + 1/(2k+1)*(2k+2)

= 1/(k+1)+1/(k+2)+…+1/(k+k) + 1/(2k+1)*(2k+2)

= 1/(k+2)+…+1/(k+k) + [ 1/(k+1)+ 1/(2k+1)*(2k+2) ]

= 1/(k+2)+…+1/(k+k) + [ 1/(k+1) ] [ 1 + 1/2 (2k+1) ]

= 1/(k+2)+…+1/(k+k) + 1/(2k+1) + 1/(2k+2)

= 1/[ (k+1)+1] + … + 1/(k+k) + 1/[(k+1)+k] + 1/[(k+1)+(k+1)] =右

等式也成立

由①②得,对于n∈N* ,等式成立