Question
For all $$n \in N,1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ..... + \frac{1}{{1 + 2 + 3 + ..... + n}}$$ is equal to
A.
$$\frac{{3n}}{{n + 1}}$$
B.
$$\frac{{n}}{{n + 1}}$$
C.
$$\frac{{2n}}{{n - 1}}$$
D.
$$\frac{{2n}}{{n + 1}}$$
Answer :
$$\frac{{2n}}{{n + 1}}$$
Solution :
Let the statement $$P\left( n \right)$$ be defined as
$$\eqalign{
& P\left( n \right):1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ..... + \frac{1}{{1 + 2 + 3 + ..... + n}} = \frac{{2n}}{{n + 1}} \cr
& {\text{i}}{\text{.e}}{\text{., }}P\left( n \right):1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ..... + \frac{2}{{n\left( {n + 1} \right)}} = \frac{{2n}}{{n + 1}} \cr} $$
Step I : For $$n = 1,P\left( 1 \right):1 = \frac{{2 \times 1}}{{1 + 1}} = \frac{2}{2} = 1,$$
which is true.
Step II : Let it is true for $$n = k,$$ i.e., $$1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ..... + \frac{2}{{k\left( {k + 1} \right)}} = \frac{{2k}}{{k + 1}}\,\,\,.....\left( {\text{i}} \right)$$
Step III : For $$n = k + 1,\left( {1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ..... + \frac{2}{{k\left( {k + 1} \right)}}} \right) + \frac{2}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$$
$$ = \frac{{2k}}{{k + 1}} + \frac{2}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$$ [using equation (i)]
$$ = \frac{{2k\left( {k + 2} \right) + 2}}{{\left( {k + 1} \right)\left( {k + 2} \right)}} = \frac{{2\left[ {{k^2} + 2k + 1} \right]}}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$$ [taking 2 common in numerator part]
$$ = \frac{{2{{\left( {k + 1} \right)}^2}}}{{\left( {k + 1} \right)\left( {k + 2} \right)}} = \frac{{2\left( {k + 1} \right)}}{{k + 2}} = \frac{{2\left( {k + 1} \right)}}{{\left( {k + 1} \right) + 1}}$$
Therefore, $$P\left( {k + 1} \right)$$ is true, when $$P\left( {k} \right)$$ is true.
Hence, from the principle of mathematical induction, the statement is true for all natural numbers $$n.$$