Question

Demonstrati ca 1 pe n( n+k) = cu 1 pe k ( 1 pe n - 1 pe n+k)

Answer 1

1/n - 1/(n+k) = (n+k)/n(n+k) - n/n(n+k) = (n+k-n)/n(n+k) =k/n(n+k)
1/k ·k/n(n+k) = 1/n(n+k)

Answer 2

$\frac{1}{n(n+k)}$=$\frac{1}{k} * \frac{k}{n(n+k)}$=

=$\frac{1}{k} * \frac{(n+k)-n}{n(n+k)}$=

=$\frac{1}{k} ( \frac{n+k}{n(n+k)} - \frac{n}{n(n+k)} )$=

=$\frac{1}{k} ( \frac{1}{n} - \frac{1}{n+k} )$


(c.c.t.d.)