Let m be a positive integer, then S=∑k=1m k1k+1k+1+1k+2+…+1m is equal to :

S=1+12+13+…+1m+212+13+…+1m+313+14+…+1m+…+(m−1)1m−1+1m+m1m=(1)(1)+12(1+2)+13(1+2+3)+14(1+2+3+4)+…+1m(1+2+3+…+m)=∑k=1m 1k(1+2+…+k)=∑k=1m 1k⋅k(k+1)2=12∑k=1m (k+1)=m4[2+(m+1)]=14m(m+3)