Problem 5 Paper For every positive integer nnn , let h(n)=1+12+13+⋯+1n.h(n) = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}.h(n)=1+21+31+⋯+n1. For example, h(1)=1h(1) = 1h(1)=1 , h(2)=1+12h(2) = 1+\frac{1}{2}h(2)=1+21 , h(3)=1+12+13h(3) = 1+\frac{1}{2}+\frac{1}{3}h(3)=1+21+31 . Prove that for n=2,3,4,…n=2,3,4,\ldotsn=2,3,4,… n+h(1)+h(2)+h(3)+⋯+h(n−1)=nh(n).n+h(1)+h(2)+h(3)+\cdots+h(n-1) = nh(n).n+h(1)+h(2)+h(3)+⋯+h(n−1)=nh(n). Answer, solutions and source