Question

If $\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\frac{1}{4\times 5\times 6}+\dots +$ $\frac{1}{100\times 101\times 102}=\frac{k}{101}$, then $34k$ is equal to

Answer 1

Answer: 286

$\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+\dots ..+\frac{1}{\mathrm{100.101.102}}=\frac{k}{101}$
$\frac{4-2}{\mathrm{2.3.4}}+\frac{5-3}{\mathrm{3.4.5}}+\dots ..+\frac{102-100}{\mathrm{100.101.102}}=\frac{2k}{101}$
$\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\dots ..+\frac{1}{100.101}-\frac{1}{101.102}=\frac{2k}{101}$
$\frac{1}{2.3}-\frac{1}{101.102}=\frac{2k}{101}$
$\therefore 2k=\frac{101}{6}-\frac{1}{102}$
$\therefore 34k=286$