a) $\frac{a}{n(n+a)}$ = $\frac{1}{n}$ - $\frac{1}{n+a}$ (n,a ∈ N*)
Ta có: $\frac{1}{n}$ - $\frac{1}{n+a}$ = $\frac{n+a}{n(n+a)}$-$\frac{n}{n(n+a)}$
= $\frac{(n+a)-n}{n(n+a)}$ = $\frac{a}{n(n+a)}$
Vậy $\frac{a}{n(n+a)}$ = $\frac{1}{n}$ - $\frac{1}{n+a}$
b) B= $\frac{5}{1.4}$ + $\frac{5}{4.7}$ +...+ $\frac{5}{100.103}$
B= $\frac{5}{3}$. ($\frac{3}{1.4}$ + $\frac{3}{4.7}$ +...+ $\frac{3}{100.103}$)
B= $\frac{5}{3}$. (1 - $\frac{1}{4}$ + $\frac{1}{4}$ - $\frac{1}{7}$ +...+ $\frac{1}{100}$ - $\frac{1}{103}$)
B= $\frac{5}{3}$.( 1- $\frac{1}{103}$ )
B= $\frac{5}{3}$. $\frac{102}{103}$
B= $\frac{170}{103}$.
Vậy B= $\frac{170}{103}$
~ Chúc bạn học tốt~
