For all $n \ge 1$,prove that: $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}$
Let $P(n)$ be the statement: $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}$.
Step $1$: For $n = 1$,the $LHS$ is $\frac{1}{1 \times 2} = \frac{1}{2}$ and the $RHS$ is $\frac{1}{1+1} = \frac{1}{2}$. Since $LHS$ = $RHS$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \ldots + \frac{1}{k(k+1)} = \frac{k}{k+1}$ $(1)$.
Step $3$: We need to prove $P(k+1)$ is true,i.e.,$\frac{1}{1 \times 2} + \ldots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2}$.
Starting from the $LHS$ of $P(k+1)$:
$= \left[ \frac{1}{1 \times 2} + \ldots + \frac{1}{k(k+1)} \right] + \frac{1}{(k+1)(k+2)}$
$= \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$ (Using $(1)$)
$= \frac{k(k+2) + 1}{(k+1)(k+2)} = \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$.
Since $P(k+1)$ is true whenever $P(k)$ is true,by the principle of mathematical induction,$P(n)$ is true for all $n \ge 1$.
Step $1$: For $n = 1$,the $LHS$ is $\frac{1}{1 \times 2} = \frac{1}{2}$ and the $RHS$ is $\frac{1}{1+1} = \frac{1}{2}$. Since $LHS$ = $RHS$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \ldots + \frac{1}{k(k+1)} = \frac{k}{k+1}$ $(1)$.
Step $3$: We need to prove $P(k+1)$ is true,i.e.,$\frac{1}{1 \times 2} + \ldots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2}$.
Starting from the $LHS$ of $P(k+1)$:
$= \left[ \frac{1}{1 \times 2} + \ldots + \frac{1}{k(k+1)} \right] + \frac{1}{(k+1)(k+2)}$
$= \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$ (Using $(1)$)
$= \frac{k(k+2) + 1}{(k+1)(k+2)} = \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$.
Since $P(k+1)$ is true whenever $P(k)$ is true,by the principle of mathematical induction,$P(n)$ is true for all $n \ge 1$.
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