Browse Questions

# For all $n \geq 1$, prove that  $\large\frac{1}{1.2}$$+\large\frac{1}{2.3}$$+\large\frac{1}{3.4}$$+...+\large\frac{1}{n(n+1)}$$=\large\frac{n}{(n+1)}$.

We can write
$P(n) : \large\frac{1}{1.2}$$+\large\frac{1}{2.3}$$+\large\frac{1}{3.4}$$+...+\large\frac{1}{n(n+1)}$$=\large\frac{n}{(n+1)}$
We note that $P(1): \large\frac{1}{1.2}$$=\large\frac{1}{2}$$ = \large\frac{1}{1+1}$, which is true. Thus, $P(n)$ is true for $n=1$.
Assume that $P(k)$ is true for some natural number $k$,
i.e., $\large\frac{1}{1.2}$$+\large\frac{1}{2.3}$$+\large\frac{1}{3.4}$$+...+\large\frac{1}{k(k+1)}$$=\large\frac{k}{(k+1)}$------------(1)
We need to prove that $P(k+1)$ is true whenever $P(k)$ is true. We have
$\large\frac{1}{1.2}$$+\large\frac{1}{2.3}$$+\large\frac{1}{3.4}$$+...+\large\frac{1}{k(k+1)}$$+\large\frac{1}{(k+1)(k+2)}$
$\bigg[ \large\frac{1}{1.2}$$+\large\frac{1}{2.3}$$+\large\frac{1}{3.4}$$+...+\large\frac{1}{k(k+1)}\bigg]$$+\large\frac{1}{(k+1)(k+2)}$