Question 23
Find $m-n$, given that $\displaystyle\frac{2}{1(1+2)}+\frac{3}{(1+2)(1+2+3)}+\frac{4}{(1+2+3)(1+2+3+4)}+\cdots+\frac{100}{(1+2+\cdots+99)(1+2+\cdots+100)}=\frac mn$.
Practice convention: m/n is in lowest terms and n is positive. The printed source does not explicitly give this condition, which is needed to make m−n unique.
Practice convention: m/n is in lowest terms and n is positive. The printed source does not explicitly give this condition, which is needed to make m−n unique.