Question

Solve the following algebraic expression for an sequential series \[3 \times 1 \times 2{\text{ }} + {\text{ }}3 \times 2 \times 3{\text{ }} + {\text{ }}3 \times 3 \times 4{\text{ }} + {\text{ }}......{\text{ }} + {\text{ }}3 \times n \times (n + 1)\]where, n is positive integral values.
(a) ${\text{(n + 1)}} \times (n + 2)$
(b) $n$
(c) $n{\text{(n + 1)}}$
(d) $n{\text{(n + 1)}}(n + 2)$

Answer 1

Hint: In the given problem, we need to require finding the generalized solution of the given sequential series which involves some certain algebraic functions. So, to solve the desire expression, we will consider the formula for both the summation of first ‘n’ natural numbers ‘N’ as well as the summation of the squares of the first ‘n’ natural numbers ‘N’. As a result, considering the difference ‘\[{t_r}\]’ of the series, by implementing the required formulae corresponding to the term, the required answer is achieved.

Complete step-by-step solution:
The given equation implies sequential series,
\[3 \times 1 \times 2{\text{ }} + {\text{ }}3 \times 2 \times 3{\text{ }} + {\text{ }}3 \times 3 \times 4{\text{ }} + {\text{ }}......{\text{ }} + {\text{ }}3 \times n \times (n + 1)\]
As a result, the series is the form of ${t_r} = 3 \times r \times (r + 1)$where, ‘r’ denotes the exact difference between terms in the series
$ \Rightarrow {t_r} = 3 \times r \times (r + 1)$
Solving the equation, we get
$
   \Rightarrow {t_r} = 3 \times ({r^2} + r) \\
   \Rightarrow {t_r} = 3{r^2} + 3r \\
 $
Now, summating the all the terms, the above equation becomes
$ \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = \sum\limits_{r = 1}^n {(3{r^2} + 3r)} $
Separating the terms,
\[ \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = \sum\limits_{r = 1}^n {3{r^2}} + \sum\limits_{r = 1}^n {3r} \]
\[ \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = 3\sum\limits_{r = 1}^n {{r^2}} + 3\sum\limits_{r = 1}^n r \]
Where, $3$is constant in the series
\[ \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = 3\left( {\sum\limits_{r = 1}^n {{r^2}} + \sum\limits_{r = 1}^n r } \right)\]… (i)
Now, since we know the standard (general) formula that, the sum of the first ‘n’ natural numbers ‘N’ is \[\sum\limits_{r = 1}^n r = \dfrac{{n(n + 1)}}{2}\], for all $n \in N$ and, that of the sum of the squares of the first ‘n’ natural numbers ‘N’ is \[\sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}\], for all $n \in N$ respectively.
Substituting these values in equation (i), we get
\[ \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = 3\left( {\dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{{n(n + 1)}}{2}} \right)\]
Solving the equation predominantly, we get
\[ \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = \dfrac{{3n(n + 1)(2n + 1)}}{6} + \dfrac{{3n(n + 1)}}{2}\]
Divide the equation by 3, we get
\[ \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = \dfrac{{n(n + 1)(2n + 1)}}{2} + \dfrac{{3n(n + 1)}}{2}\]
As denominator is same, we can add both the terms, we get
\[ \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = \dfrac{1}{2}\left[ {n(n + 1)(2n + 1) + 3n(n + 1)} \right]\]
Solving the equations mathematically, we get
\[
   \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = \dfrac{{n(n + 1)}}{2}\left[ {2n + 1 + 3} \right] \\
   \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = \dfrac{{n(n + 1)}}{2}\left[ {2n + 4} \right] \\
 \]
Again, dividing the bracket by $2$, we get
\[ \Rightarrow \sum\limits_{r = 1}^n {{t_r}} = n(n + 1)(n + 2)\]
Hence, the required solution is obtained!
$\therefore $ Option (d) is correct!

Note: For finding the solution of the sequential series problem algebraically, we have to first calculate the ‘\[{t_r}\]’ i.e. the exact difference between the series here the difference is, ${t_r} = 3{r^2} + 3r$. As a result, to that term we have the generalized formulae for both the summation of squares as well as for all real numbers i.e. \[\sum\limits_{r = 1}^n r = \dfrac{{n(n + 1)}}{2}\]and\[\sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}\] respectively. These formulae are also known as Sigma-equations. Learning these equations or formulae will really helpful to solve these drastic problem as per as sequential series is concerned. Care should be taken while solving the certain equations.