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 $