Question

Find the sum: \[\dfrac{3}{{{1^2} \cdot {2^2}}} + \dfrac{5}{{{2^2} \cdot {3^2}}} + \dfrac{7}{{{3^2} \cdot {4^2}}} + ....n{\rm{ terms}}\]

Answer 1

Hint:
Here we have to find the sum of the series. For that, we will first break the numerator of each term into the difference of the squares of the two numbers given exactly in the denominator. Then we will find the \[{r^{th}}\] term of the series. We will simplify the \[{r^{th}}\]term further and we will write all the terms in the same form as the\[{r^{th}}\]term, then we will find the sum of all the terms and we will get the required sum of the series.

Complete step by step solution:
Let’s first consider the first term of the series which is \[\dfrac{3}{{{1^2} \cdot {2^2}}}\].
We can write the numerator of \[\dfrac{3}{{{1^2} \cdot {2^2}}}\] as \[{2^2} - {1^2}\].
So the first term becomes \[\dfrac{{{2^2} - {1^2}}}{{{1^2} \cdot {2^2}}}\].
Similarly, we can write second term of the series as \[\dfrac{{{3^2} - {2^2}}}{{{2^2} \cdot {3^2}}}\]
We can write second term of the series as \[\dfrac{{{4^2} - {3^2}}}{{{4^2} \cdot {3^2}}}\]and so on.
Thus, we can write the \[{r^{th}}\] of a series as \[{T_r} = \dfrac{{{r^2} - {{\left( {r - 1} \right)}^2}}}{{{r^2} \cdot {{\left( {r - 1} \right)}^2}}}\].
We will now simplify the \[{r^{th}}\] term of the series.
\[{T_r} = \dfrac{{{r^2}}}{{{{\left( {r - 1} \right)}^2} \cdot {r^2}}} - \dfrac{{{{\left( {r - 1} \right)}^2}}}{{{{\left( {r - 1} \right)}^2} \cdot {r^2}}}\]
On further simplification, we get
\[{T_r} = \dfrac{1}{{{{\left( {r - 1} \right)}^2}}} - \dfrac{1}{{{r^2}}}\]
Thus, the sum of the series \[{S_n} = \sum\limits_{r = 2}^n {{T_r}} \] .
Putting value of\[{T_r}\] in above equation, we get
\[{S_n} = \sum\limits_{r = 2}^n {\dfrac{1}{{{{\left( {r - 1} \right)}^2}}} - \dfrac{1}{{{r^2}}}} \]
Thus on expansion, we get
\[{S_n} = \dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}} + .... + \dfrac{1}{{{{\left( {n - 1} \right)}^2}}} - \dfrac{1}{{{n^2}}}\]
On subtraction of the similar terms, the sum becomes
\[{S_n} = 1 - \dfrac{1}{{{n^2}}}\]
We can also write the sum as
\[{S_n} = \dfrac{{{n^2} - 1}}{{{n^2}}}\]

Hence, the sum of the given expression is \[\dfrac{{{n^2} - 1}}{{{n^2}}}\].

Note:
Since we have calculated the sum of the series, following points should be remembered while solving this problems:-
1) A ‘sum of a series’ is defined as the result obtained after addition of each term of a series.
2) Example- \[1 + 2 + 3 + 4\] is a series but 10 is the sum of the required series, which is the result of addition of each term of the series.