Question

Find sum of the series\[\dfrac{1}{{1 \times 3 \times 5}} + \dfrac{1}{{3 \times 5 \times 7}} + \dfrac{1}{{5 \times 7 \times 9}}......\]?

Answer 1

Hint: For finding the sum of a series first you have to know that the given expression is in which series, that is it is in arithmetic progression or geometric progression or some mixed series. To know about the series you have to find the common difference of the series and accordingly you can go through the question after identifying the series.

Complete step by step solution:
The given series is \[\dfrac{1}{{1 \times 3 \times 5}} + \dfrac{1}{{3 \times 5 \times 7}} + \dfrac{1}{{5 \times 7 \times 9}}......\]
Here no common factors are getting out, now we have to solve the series by general summation rule.
Let’s modify our given expression by including variable “x” such that the value of “x” is one. Now writing the equation in form of variable and introducing summation sign we get:
\[ \Rightarrow 1 = 2x - 1,\,3 = 2x + 1,\,5 = 2x + 3\]
Introducing this expression in the first term of the expression we get:
\[
   \Rightarrow \sum\limits_{x = 1}^x {\dfrac{1}{{(2x - 1)(2x + 1)(2x + 3)}}} \\
  multiplying\,and\,dividing\,by\,4\,in\,numerator\,and\,deno\min ator\,we\,get \\
   \Rightarrow \dfrac{1}{4}\sum\limits_{x = 1}^x {\dfrac{4}{{(2x - 1)(2x + 1)(2x + 3)}}} \\
   \Rightarrow \dfrac{1}{4}\sum\limits_{x = 1}^x {\dfrac{{(2x + 3) - (2x - 1)}}{{(2x - 1)(2x + 1)(2x + 3)}}} \\
   \Rightarrow \dfrac{1}{4}\sum\limits_{x = 1}^x {\left[ {\dfrac{{2x + 3}}{{(2x - 1)(2x + 1)(2x + 3)}} - \dfrac{{2x - 1}}{{(2x - 1)(2x + 1)(2x + 3)}}} \right]} \\
 \]
\[
   \Rightarrow \dfrac{1}{4}\sum\limits_{x = 1}^x {\left[ {\dfrac{1}{{(2x - 1)(2x + 1)}} - \dfrac{1}{{(2x + 1)(2x + 3)}}} \right]} \\
  we\,know\,2x - 1 = 1,\,and\,2x + 1 = 3\,putting\,this\,in\,above\,\exp ression\,we\,get: \\
   \Rightarrow \dfrac{1}{4}\sum\limits_{x = 1}^x {\left[ {\dfrac{1}{{1 \times 3}} - \dfrac{1}{{(2x + 1)(2x + 3)}}} \right]} \\
 here\,we\,can\,remove\,the\,summation\,sign\,and\,only\,two\,term\,will\,remain\,here\,rest\,will\,cancel\,out\,and\,become\,zero. \\
   \Rightarrow \dfrac{1}{4}\left[ {\dfrac{1}{{1 \times 3}} - \dfrac{1}{{(2x + 1)(2x + 3)}}} \right] \\
 \]\[ \Rightarrow \left[ {\dfrac{{(2x + 1)(2x + 3) - 3}}{{4 \times 1 \times 3 \times (2x + 1)(2x + 3)}}} \right]\]
\[ \Rightarrow \left[ {\dfrac{{4{x^2} + 8x + 3 - 3}}{{4 \times 1 \times 3 \times (2x + 1)(2x + 3)}}} \right]\]
\[
   \Rightarrow \left[ {\dfrac{{4({x^2} + 2x)}}{{4 \times 1 \times 3 \times (2x + 1)(2x + 3)}}} \right] \\
   \Rightarrow \left[ {\dfrac{{({x^2} + 2x)}}{{3 \times (2x + 1)(2x + 3)}}} \right] \\
   \Rightarrow \left[ {\dfrac{{x(x + 2)}}{{3 \times (4{x^2} + 8x + 3)}}} \right] \;
 \]
This is the required result that is summation of the given expression, here after putting the value of “x” we get:
\[ \Rightarrow \left[ {\dfrac{{1(1 + 2)}}{{3 \times (4{{(1)}^2} + 8(1) + 3)}}} \right]\]
\[
   \Rightarrow \left[ {\dfrac{3}{{45}}} \right] \\
   \Rightarrow \dfrac{1}{{15}} \;
 \]
This is our final result of the expression.
So, the correct answer is “$\dfrac{1}{{15}}$”.

Note: Here we have broken the term by introducing variables and then we get that terms are getting cancelled as repetition is there, and then we get our result.
For finding the summation of the series you have to deal with the solution like break the term in such a way that terms get canceled out as repetition is there with the terms, and if not possible then you have to check for the general identity and then use the standard formulae accordingly.