Browse Questions

# Prove the following by using the principle of mathematical induction for all $n \in N$  $1+ \large\frac{1}{(1+2)}$$+ \large\frac{1}{(1+2+3)}$$+...+ \large\frac{1}{(1+2+3...n)}$$= \large\frac{2n}{(n+1)} Can you answer this question? ## 1 Answer 0 votes Let the given statement be P(n), i.e., P(n) : 1+ \large\frac{1}{1+2}$$+\large\frac{1}{1+2+3}$$+...+ \large\frac{1}{1+2+3+...n}$$ = \large\frac{2n}{n+1}$
For $n=1$, we have
$P(1) : 1 = \large\frac{2.1}{1+1}$$= \large\frac{2}{2}$$ =1$, which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$= 1+ \large\frac{1}{1+2}$$+\large\frac{1}{1+2+3}$$ +...+ \large\frac{1}{1+2+3+...+k} $$= \large\frac{2k}{k+1}---------(i) We shall now prove that P(k+1) is true. Consider = 1+ \large\frac{1}{1+2}$$+\large\frac{1}{1+2+3}$$+...+ \large\frac{1}{1+2+3+...+k}$$+\large\frac{1}{1+2+3+...+k+(k+1)}$
$=\bigg( 1+ \large\frac{1}{1+2}$$+\large\frac{1}{1+2+3}$$ +...+ \large\frac{1}{1+2+3+...+k}\bigg) $$+\large\frac{1}{1+2+3+...+k+(k+1)} = \large\frac{2k}{k+1}$$+ \large\frac{1}{1+2+3+...+k+(k+1)} \qquad$ [ Using (i) ]
$= \large\frac{2k}{k+1}$$+ \large\frac{1}{ \bigg( \Large\frac{(k+1)(k+1+1)}{2} \bigg)}$$ \qquad \bigg[1+2+3+..+n = \large\frac{n(n+1)}{2} \bigg]$
$= \large\frac{2k}{k+1} $$+ \large\frac{2}{(k+1)(k+2)} \large\frac{2}{(k+1)}$$ \bigg( k+ \large\frac{1}{k+2}\bigg)$