Question

Prove that $lo{g}_{e}a=1+\frac{2}{\mathrm{1.2.3}}+\frac{2}{\mathrm{3.4.5}}+\frac{2}{\mathrm{5.6.7}}+...$ Find a

Answer 1

$lo{g}_{e}a=1+\frac{2}{\mathrm{1.2.3}}+\frac{2}{\mathrm{3.4.5}}+\frac{2}{\mathrm{5.6.7}}+...$
Here,$\frac{2}{\mathrm{1.2.3}}=\frac{1}{1.2}-\frac{1}{2.3}$
Similarly, $\frac{2}{\mathrm{3.4.5}}=\frac{1}{3.4}-\frac{1}{4.5}$
So $S=1+(\frac{1}{1.2}-\frac{1}{2.3})+(\frac{1}{3.4}-\frac{1}{4.5})+.....$
$\Rightarrow S=1+(\frac{1}{1.2}+\frac{1}{3.4}+....)-(\frac{1}{2.3}+\frac{1}{4.5}+....)$
$\Rightarrow S=1+(\frac{1}{1.2}+\frac{1}{{3.2}^2}+....)-(\frac{1}{2.3}+\frac{1}{4.5}+....)$
$\Rightarrow S=1+{\log }_{e}2-[(\frac{1}{2}-\frac{1}{3})+(\frac{1}{4}-\frac{1}{5})+.....]$
$\Rightarrow S=1+{\log }_{e}2-1+{\log }_{e}2$
$\Rightarrow S=2{\log }_{e}2={\log }_{e}4$
Comparing with given value, we get
$a=4$