n=k時,最后一項(xiàng)是1/(2^k-1)
n=k+1時,最后一項(xiàng)是1/[2^(k+1)-1]
所以增加的是從1/2^k到1/[2^(k+1)-1]
一共2^(k+1)-1-2^k+1
=2*2^k-2^k
=2^k項(xiàng)
n=1成立
設(shè)n=k成立,k>=1
n=k+1
1*2+2*3+……+n(n+1)+(n+1)(n+2)=n(n+1)(n+2)/3+(n+1)(n+2)
=(n+1)(n+2)[n/3+1]
=(n+1)(n+2)(n+3)/3
=[(n+1)][(n+1)+1][(n+1)+2]/3
綜合以上,得證