The sum of the first 4 terms of \(\dfrac{1}{4\times 8} + \dfrac{1}{8\times 12} + \dfrac{1}{12\times 16}+\)_____ is:

Calculation:

\(\dfrac{1}{4\times 8} + \dfrac{1}{8\times 12} + \dfrac{1}{12\times 16}\)

\(\dfrac{1}{4\times 8} + {(\dfrac{1}{1} + \dfrac{1}{3}+\dfrac{1}{6} +\dfrac{1}{10} )}\)

\(\dfrac{1}{4\times 8} {(\dfrac{1}{1} + \dfrac{1}{3 \times 1}+\dfrac{1}{3 \times 2}+\dfrac{1}{2 \times 5})}\)

\(\dfrac{1}{4\times 8} {(\dfrac{96}{60})} = 1/20\)

Alternate Method

\(\dfrac{1}{4\times 8} + \dfrac{1}{8\times 12} + \dfrac{1}{12\times 16} + ...\)

Each term is in the form \(1 \over {4n \times 4(n + 1)}\)

⇒ for the fourth term n = 4

⇒ The fourth term should be \({1 \over {(4 \times 4) \times [4 \times (4 + 1)}]}= {1 \over {16 \times 20}}\)

The series will become \(\dfrac{1}{4\times 8} + \dfrac{1}{8\times 12} + \dfrac{1}{12\times 16} + \dfrac{1}{16\times 20}\)

⇒ \({1 \over 4}[({1 \over 4} - {1 \over 8}) + ({1 \over 8} - {1 \over 12}) + ({1 \over 12} - {1 \over 16}) + ({1 \over 16} - {1 \over 20})]\)

⇒ \({1 \over 4} \times ({1 \over 4} - {1 \over 20})\)

⇒ \({1 \over 4} \times {4 \over 20}\)

⇒ 1/20

∴ The required sum of the first four terms of the series is 1/20.