Question

The sum to n terms of the series:
$1.3.5 + 3.5.7 + 5.7.9 + \ldots$

Answer 1

Hint : The terms of the series consist of three consecutive odd numbers. The formula to be used for the three consecutive odd integers is $\left( {2n - 1} \right)$ , $\left( {2n + 1} \right)$ , and $\left( {2n + 3} \right)$ respectively. The n-th term of the series is formed and after that its summation is done to obtain the sum of the series for the n terms.

Complete step-by-step answer :
The given series is
$1.3.5 + 3.5.7 + 5.7.9 + \ldots$
Each term of the series should be observed carefully, in order to obtain the n-th term of the series.
The first terms include the product of $1$ , $3$ and $5$ . The three terms are three consecutive odd numbers. In which the first term starts with $1$ and ends with $5$ . The successive number in each term differs by $2$ .
The second term includes the product of $3$ , $5$ and $7$ . The three terms are three consecutive odd numbers and the first term of it starts with $3$ and ends with $7$ . The successive number in each term differs by $2$ .
After observing the two terms, the n-th term of the series should be suitably chosen.
Let us assume the n-th term of the series is $\left( {2n - 1} \right)$ , $\left( {2n + 1} \right)$ and $\left( {2n + 3} \right)$ .
If we put $n = 1$ we get
$\left( {2 \times 1 - 1} \right),\left( {2 \times 1 + 1} \right)$ and $\left( {2 \times 1 + 3} \right)$ or $1,3$ and $5$ respectively.
If we put $n = 2$ we get
$3,5$ and $7$ .
The n-th term of the series is given by
${T_n} = \left( {2n - 1} \right)\left( {2n + 1} \right)\left( {2n + 3} \right) \cdots \left( 1 \right)$
Taking the summation of the n-th term in equation (1),

$${S_n} = \sum\limits_{n = 1}^n {{T_n}} \\\ {S_n} = \sum\limits_{n = 1}^n {\left( {2n - 1} \right)\left( {2n + 1} \right)\left( {2n + 3} \right) \cdots \left( 2 \right)} \\\$$

Now opening the brackets in equation (2), we get

$$\Rightarrow {S_n} = \sum\limits_{n = 1}^n {\left( {2n - 1} \right)\left( {2n + 1} \right)\left( {2n + 3} \right)} \\\ {S_n} = \sum\limits_{n = 1}^n {\left( {4{n^2} - 1} \right)\left( {2n + 3} \right)} \\\ \Rightarrow {S_n} = \sum\limits_{n = 1}^n {\left( {8{n^3} + 12{n^2} - 2n - 3} \right)} \cdots \left( 3 \right) \;$$

Now taking the summation of the individual terms

$$\Rightarrow {S_n} = \sum\limits_{n = 1}^n {\left( {8{n^3}} \right) + } \sum\limits_{n = 1}^n {\left( {12{n^2}} \right) + } \sum\limits_{n = 1}^n {\left( { - 2n} \right) + \sum\limits_{n = 1}^n {\left( { - 3} \right)} } \\\ {S_n} = 8\sum\limits_{n = 1}^n {\left( {{n^3}} \right) + } 12\sum\limits_{n = 1}^n {\left( {{n^2}} \right) - 2} \sum\limits_{n = 1}^n {\left( n \right) - 3\sum\limits_{n = 1}^n {\left( 1 \right)} } \cdots \left( 4 \right) \;$$

The sum of the n-th of the series is
$\sum {{n^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{{{2^2}}}$ , $\sum {{n^2} = \dfrac{{\left( n \right)\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}$ , $\sum n = \dfrac{{\left( n \right)\left( {n + 1} \right)}}{2}$ and $\sum 1 = n$ .
Substitute the values in equation (4), we get
$\Rightarrow {S_n} = 8 \times \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{{{2^2}}} + 12 \times \dfrac{{\left( n \right)\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - 2 \times \dfrac{{\left( n \right)\left( {n + 1} \right)}}{2} - 3n \\\ {S_n} = 2 \times {n^2}{\left( {n + 1} \right)^2} + 2 \times \left( n \right)\left( {n + 1} \right)\left( {2n + 1} \right) - \left( n \right)\left( {n + 1} \right) - 3n \cdots \left( 5 \right) \;$
Taking out common from the first three terms in equation (5) we get
$\Rightarrow {S_n} = n\left( {n + 1} \right)\left[ {2n\left( {n + 1} \right) + 2\left( {2n + 1} \right) - 1} \right] - 3n \\\ \Rightarrow {S_n} = n\left( {n + 1} \right)\left[ {2{n^2} + 2n + 4n + 2 - 1} \right] - 3n \\\ \Rightarrow {S_n} = n\left( {n + 1} \right)\left[ {2{n^2} + 6n + 1} \right] - 3n \;$
Hence, the sum of the series is ${S_n} = n\left( {n + 1} \right)\left[ {2{n^2} + 6n + 1} \right] - 3n$ .
So, the correct answer is “ $n\left( {n + 1} \right)\left[ {2{n^2} + 6n + 1} \right] - 3n$ ”.

Note : The important thing is to analyze each and every term. After observing the pattern the n-the term should be decided
For consecutive odd terms, $\left( {2n - 1} \right),\left( {2n + 1} \right)$ and $\left( {2n + 3} \right)$ can be chosen.
For even terms $\left( {2n} \right),\left( {2n + 2} \right)$ and $\left( {2n + 4} \right)$ can be chosen.
The following summations are important and should be remembered.
$\sum {{n^3} = } \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{{{2^2}}}$
$\sum {{n^2} = } \dfrac{{\left( n \right)\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
$\sum n = \dfrac{{n\left( {n + 1} \right)}}{2}$
$\sum 1 = n$