Question

Find the sum of series 1/1.2.3.4 + 1/2.3.4.5 + 1/3.4.5.6

Answer 1

Answer:

$\frac{1}{18} - \frac{1}{3.(n+1)(n+2).(n+3)}$

Step-by-step explanation:

Find the sum of series 1/1.2.3.4 + 1/2.3.4.5 + 1/3.4.5.6

Nth Term is

=$\frac{1}{n.(n+1)(n+2)(n+3)} \\ \\= \frac{1}{n.(n+1)} \times \frac{1}{(n+2).(n+3)} \\\\= (\frac{1}{n} - \frac{1}{(n+1)} ) ( \frac{1}{(n+2)} - \frac{1}{(n+3)})$

$= \frac{1}{n}.\frac{1}{(n+2)} - \frac{1}{n}.\frac{1}{(n+3)} - \frac{1}{(n+1)}.\frac{1}{(n+2)} + \frac{1}{(n+1)}\frac{1}{(n+3)}\\ \\= (\frac{1}{2}) (\frac{1}{n} - \frac{1}{(n+2)}) - (\frac{1}{3}) (\frac{1}{n} - \frac{1}{(n+3)}) - (\frac{1}{n+1} - \frac{1}{(n+2)}) + (\frac{1}{2}) (\frac{1}{n+1} - \frac{1}{(n+3)})$

$= (\frac{1}{6}) (\frac{1}{n}) - (\frac{1}{2}) (\frac{1}{n+1}) + (\frac{1}{2}) (\frac{1}{n+2}) - (\frac{1}{6}) (\frac{1}{n+3})$

This can be used to find Sum

or

$\frac{1}{1.2.3.4} = (\frac{1}{3})(\frac{1}{1.2.3}- \frac{1}{2.3.4})\\ \\\frac{1}{2.3.4.5} = (\frac{1}{3})(\frac{1}{2.3.4}- \frac{1}{3.4.5}) \\ \\\frac{1}{3.4.5.6} = (\frac{1}{3})(\frac{1}{3.4.5} - \frac{1}{4.5.6}) \\ \\......\\......\\\frac{1}{n.(n+1).(n+2)(n+3)} = (\frac{1}{3})(\frac{1}{(n)(n+1)(n+2)} - \frac{1}{(n+1)(n+2).(n+3)})$

Adding all we get

$\frac{1}{3} (\frac{1}{1.2.3} - \frac{1}{(n+1)(n+2).(n+3)}) \\ \\\frac{1}{18} - \frac{1}{3.(n+1)(n+2).(n+3)}$

Answer 2

Answer:

refer to the attachment!!