8.0 Miscellaneous Series

This method is used to find the $n$th term and sum of the series in which the differences of the successive terms are in A.P. or G.P.

The method is explained with the help of example.

Question 18. Find the $n$th term of the series $$1 + 2 + 5 + 12 + 25 + 46 + ...$$

Solution: Clearly here the differences between the successive terms are $$\begin{equation} \begin{aligned} 2 - 1,5 - 2,12 - 5,25 - 12,46 - 25,... \\ 1,3,7,13,21,... \\\end{aligned} \end{equation} $$

which are neither in A.P. nor in G.P. Therefore, we can not use the method of difference to find the $n$th term but the ${2^{nd}}$ difference between the successive terms are $$\begin{equation} \begin{aligned} 3 - 1,7 - 3,13 - 7,21 - 13,... \\ 2,4,6,8,... \\\end{aligned} \end{equation} $$

which are in A.P. Therefore, now we can use the method of difference to find the $n$th term.

First of all, write the sum of the series as shown.

$$\begin{equation} \begin{aligned} {S_n} = 1 + 2 + 5 + 12 + 25 + 46 + ... + {T_{n - 1}} + {T_n}\quad ...(1) \\ {S_n} = {\text{ 1 + 2 + 5 + 12 + 25 + }}.............{\text{, + }}{T_{n - 1}} + {T_n}\quad ...(2) \\\end{aligned} \end{equation} $$

Subtracting $(2)$ from $(1)$, we get

$$\begin{equation} \begin{aligned} 0 = 1 + 1 + 3 + 7 + 13 + 21 + ... + ({T_n} - {T_{n - 1}} - {T_n}) \\ {T_n} = 1 + 1 + 3 + 7 + 13 + 21 + ... + {t_{n - 1}} + {t_n}\quad ...(3) \\\end{aligned} \end{equation} $$

(Here $n$th term of ${T_n}$ is ${t_n}$)

Now, rewrite the ${T_n}$ as shown.

$$\begin{equation} \begin{aligned} {T_n} = 1 + 1 + 3 + 7 + 13 + 21 + ... + {t_{n - 1}} + {t_n} \\ {T_n} = {\text{ 1 + 1 + 3 + 7 + 13 + 21 + }}......{\text{ + }}{t_{n - 1}} + {t_n}\quad ...(4) \\\end{aligned} \end{equation} $$

Subtracting $(4)$ from $(3)$, we get

$$\begin{equation} \begin{aligned} 0 = 1 + 0 + 2 + 4 + 6 + 8 + ... + ({t_n} - {t_{n - 1}}) - {t_n} \\ {t_n} = 1 + 2 + 4 + 6 + 8 + ... + (n - 1){\text{ terms}} \\ {\text{ = 1 + (2 + 4 + 6 + 8 + }}...{\text{ + }}(n - 2){\text{ terms)}} \\ {\text{ = 1 + }}\frac{{n - 2}}{2}\left[ {2.2 + (n - 2 - 1)2} \right] \\ {t_n}{\text{ = }}{n^2} - 3n + 3 \\\end{aligned} \end{equation} $$

Now, we have to find ${T_n}$ i.e., $$\begin{equation} \begin{aligned} {T_n} = \sum {{t_n}} = \sum {{n^2} - 3\sum n + 3\sum 1 } \\ {\text{ = }}\frac{{n(n + 1)(2n + 1)}}{6} - \frac{{3n(n + 1)}}{2} + \frac{{3n}}{1} \\ {\text{ = }}\frac{n}{6}\left( {2{n^2} - 6n + 10} \right) \\ {\text{ = }}\frac{n}{3}\left( {{n^2} - 3n + 5} \right) \\\end{aligned} \end{equation} $$

This method is used to find the sum of the series of the following type: $$\begin{equation} \begin{aligned} {a_1}{a_2}...{a_r} + {a_2}{a_3}...{a_{r + 1}} + ... + {a_n}{a_{n + 1}}...{a_{n + r - 1}} \\ {\text{or}} \\ \frac{1}{{{a_1}{a_2}...{a_r}}} + \frac{1}{{{a_2}{a_3}...{a_{r + 1}}}} + ... + \frac{1}{{{a_n}{a_{n + 1}}...{a_{n + r - 1}}}} \\\end{aligned} \end{equation} $$

where ${a_1},{a_2},{a_3}...{a_n}...$ are in A.P.

The method is explained with the help of example.

Question 19. Find the sum of series $$\frac{1}{{1.2.3.4}} + \frac{1}{{2.3.4.5}} + ...n{\text{ terms}}$$

Solution: Since $1,2,3,4...$ are in A.P., we can use ${V_r}$ method to find the sum of given series.

Let general term of the series is $${T_r} = \frac{1}{{r(r + 1)(r + 2)(r + 3)}}$$

Multiply and divide by the difference of first and last term in the general term of series i.e.,

$$\begin{equation} \begin{aligned} {T_r} = \frac{{r + 3 - r}}{{r(r + 1)(r + 2)(r + 3)}} \times \frac{1}{{r + 3 - r}} \\ {\text{ = }}\frac{{r + 3 - r}}{{3r(r + 1)(r + 2)(r + 3)}} \\ {\text{ = }}\frac{1}{3}\left[ {\frac{1}{{r(r + 1)(r + 2)}} - \frac{1}{{(r + 1)(r + 2)(r + 3)}}} \right] \\\end{aligned} \end{equation} $$

Therefore, $$\begin{equation} \begin{aligned} {T_r}{\text{ = }}\frac{1}{3}\left[ {\frac{1}{{r(r + 1)(r + 2)}} - \frac{1}{{(r + 1)(r + 2)(r + 3)}}} \right] \\ {\text{ = }}\frac{1}{3}\left[ {{V_{r - 1}} - {V_r}} \right] \\\end{aligned} \end{equation} $$

Now, Sum of the series is $$\begin{equation} \begin{aligned} {S_r} = \sum\limits_{r = 1}^n {{T_r}} = \frac{1}{3}\sum\limits_{r = 1}^n {\left[ {{V_{r - 1}} - {V_r}} \right]} = \frac{1}{3}\left[ {{V_0} + {V_1} + {V_2} + ... + {V_{n - 1}} - ({V_1} + {V_2} + ... + {V_{n - 1}} + {V_n})} \right] \\ {\text{ = }}\frac{1}{3}\left( {{V_0} - {V_n}} \right) \\ {\text{ = }}\frac{1}{3}\left( {\frac{1}{{1.2.3}} - \frac{1}{{(n + 1)(n + 2)(n + 3)}}} \right) \\\end{aligned} \end{equation} $$