QUESTION

# The sum of the series $\dfrac{1}{{3 \times 7}} + \dfrac{1}{{7 \times 11}} + \dfrac{1}{{11 \times 15}} + ............$is$a.{\text{ }}\dfrac{1}{3} \\ b.{\text{ }}\dfrac{1}{6} \\ c.{\text{ }}\dfrac{1}{9} \\ d.{\text{ }}\dfrac{1}{{12}} \\$

Hint:First calculate the general term of the given series i.e. its ${n^{th}}$ term. Using the formula ${a_n} = {a_1} + \left( {n - 1} \right)d$, where d is the common difference. In ${n^{th}}$ term numerator remains the same which is 1, but the denominator will follow the property of A.P.

As we see that $\left( {3,7,11...........} \right)$forms an A.P
With first term ${a_1} = 3$, common difference $d = \left( {7 - 3} \right) = \left( {11 - 7} \right) = 4$, and the number of terms is $n$.

Therefore the ${n^{th}}$term of the series is written as
${a_n} = {a_1} + \left( {n - 1} \right)d \\ \Rightarrow {a_n} = 3 + \left( {n - 1} \right)4 = 4n - 1 \\$

Now as we see that $\left( {7,11,15...........} \right)$ forms an A.P

With first term${a_1} = 7$, common difference $d = \left( {11 - 7} \right) = \left( {15 - 11} \right) = 4$, and the number of terms is $n$.

Therefore the ${n^{th}}$ term of the series is written as
${a_n} = {a_1} + \left( {n - 1} \right)d \\ \Rightarrow {a_n} = 7 + \left( {n - 1} \right)4 = 4n + 3 \\$

Therefore the ${n^{th}}$ term of the given series is ${T_n} = \dfrac{1}{{\left( {4n - 1} \right)\left( {4n + 3} \right)}}$
$\Rightarrow {T_n} = \dfrac{1}{{\left( {4n - 1} \right)\left( {4n + 3} \right)}} = \dfrac{1}{4}\left[ {\dfrac{1}{{4n - 1}} - \dfrac{1}{{4n + 3}}} \right]$

Now for $n = 1,2,3...........,\left( {n - 1} \right),n$
$\Rightarrow {T_1} = \dfrac{1}{4}\left[ {\dfrac{1}{3} - \dfrac{1}{7}} \right] \\ {\text{ }}{T_2} = \dfrac{1}{4}\left[ {\dfrac{1}{7} - \dfrac{1}{{11}}} \right] \\ {\text{ }}{T_3} = \dfrac{1}{4}\left[ {\dfrac{1}{{11}} - \dfrac{1}{{15}}} \right] \\ {\text{ }}{\text{.}} \\ {\text{ }}{\text{.}} \\ {\text{ }}{T_{n - 1}} = \dfrac{1}{4}\left[ {\dfrac{1}{{4\left( {n - 1} \right) - 1}} - \dfrac{1}{{4\left( {n - 1} \right) + 3}}} \right] \\ {\text{ }}{T_n} = \dfrac{1}{4}\left[ {\dfrac{1}{{4n - 1}} - \dfrac{1}{{4n + 3}}} \right] \\$

Now add all L.H.S and R.H.S respectively
$\Rightarrow {T_1} + {T_2} + {T_3} + ....... + {T_{n - 1}} + {T_n} = \dfrac{1}{4}\left[ {\dfrac{1}{3} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{11}} + \dfrac{1}{{11}} - \dfrac{1}{{15}} + ............ + \dfrac{1}{{4\left( {n - 1} \right) - 1}} - \dfrac{1}{{4\left( {n - 1} \right) + 3}} + \dfrac{1}{{4n - 1}} - \dfrac{1}{{4n + 3}}} \right] \\ \Rightarrow {T_1} + {T_2} + {T_3} + ....... + {T_{n - 1}} + {T_n} = \dfrac{1}{4}\left[ {\dfrac{1}{3} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{11}} + \dfrac{1}{{11}} - \dfrac{1}{{15}} + ............ + \dfrac{1}{{4\left( {n - 1} \right) - 1}} - \dfrac{1}{{4n - 1}} + \dfrac{1}{{4n - 1}} - \dfrac{1}{{4n + 3}}} \right] \\ \Rightarrow \sum\limits_{r = 1}^n {{T_r} = } \dfrac{1}{4}\left[ {\dfrac{1}{3} - \dfrac{1}{{4n + 3}}} \right] \\$

Now, if the series tends to infinite i.e., $n = \infty$
$\Rightarrow \sum\limits_{r = 1}^n {{T_r} = } \dfrac{1}{4}\left[ {\dfrac{1}{3} - \dfrac{1}{{4n + 3}}} \right] = \dfrac{1}{4}\left[ {\dfrac{1}{3} - \dfrac{1}{{\infty + 3}}} \right] = \dfrac{1}{4}\left[ {\dfrac{1}{3} - \dfrac{1}{\infty }} \right]$

As we know $\dfrac{1}{\infty } = 0$
$\Rightarrow \sum\limits_{r = 1}^n {{T_r} = } \dfrac{1}{4}\left[ {\dfrac{1}{3} - \dfrac{1}{{4n + 3}}} \right] = \dfrac{1}{4}\left[ {\dfrac{1}{3} - \dfrac{1}{\infty }} \right] = \dfrac{1}{4}\left[ {\dfrac{1}{3} - 0} \right] = \dfrac{1}{{12}}$

So, this is the required sum of the given series.

Hence, option (d) is correct.

Note: In such types of questions the key concept we have to remember is that first always find out its ${n^{th}}$ term, using some basic properties of A.P which is stated above, then calculate its term for $n = 1,2,3...........,\left( {n - 1} \right),n$, then add all the L.H.S and R.H.S respectively, we will get the required sum of the given series, then substitute n equals to infinity, we will get the required answer.