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Question: Find the sum: \[\dfrac{3}{{{1^2} \cdot {2^2}}} + \dfrac{5}{{{2^2} \cdot {3^2}}} + \dfrac{7}{{{3^2} \...

Find the sum: 312⋅22+522⋅32+732⋅42+....nterms\dfrac{3}{{{1^2} \cdot {2^2}}} + \dfrac{5}{{{2^2} \cdot {3^2}}} + \dfrac{7}{{{3^2} \cdot {4^2}}} + ....n{\rm{ terms}}

Explanation

Solution

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 rth{r^{th}} term of the series. We will simplify the rth{r^{th}}term further and we will write all the terms in the same form as therth{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 312⋅22\dfrac{3}{{{1^2} \cdot {2^2}}}.
We can write the numerator of 312⋅22\dfrac{3}{{{1^2} \cdot {2^2}}} as 22−12{2^2} - {1^2}.
So the first term becomes 22−1212⋅22\dfrac{{{2^2} - {1^2}}}{{{1^2} \cdot {2^2}}}.
Similarly, we can write second term of the series as 32−2222⋅32\dfrac{{{3^2} - {2^2}}}{{{2^2} \cdot {3^2}}}
We can write second term of the series as 42−3242⋅32\dfrac{{{4^2} - {3^2}}}{{{4^2} \cdot {3^2}}}and so on.
Thus, we can write the rth{r^{th}} of a series as Tr=r2−(r−1)2r2⋅(r−1)2{T_r} = \dfrac{{{r^2} - {{\left( {r - 1} \right)}^2}}}{{{r^2} \cdot {{\left( {r - 1} \right)}^2}}}.
We will now simplify the rth{r^{th}} term of the series.
Tr=r2(r−1)2⋅r2−(r−1)2(r−1)2⋅r2{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
Tr=1(r−1)2−1r2{T_r} = \dfrac{1}{{{{\left( {r - 1} \right)}^2}}} - \dfrac{1}{{{r^2}}}
Thus, the sum of the series Sn=∑r=2nTr{S_n} = \sum\limits_{r = 2}^n {{T_r}} .
Putting value ofTr{T_r} in above equation, we get
Sn=∑r=2n1(r−1)2−1r2{S_n} = \sum\limits_{r = 2}^n {\dfrac{1}{{{{\left( {r - 1} \right)}^2}}} - \dfrac{1}{{{r^2}}}}
Thus on expansion, we get
Sn=112−122+122−132+132−142+....+1(n−1)2−1n2{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
Sn=1−1n2{S_n} = 1 - \dfrac{1}{{{n^2}}}
We can also write the sum as
Sn=n2−1n2{S_n} = \dfrac{{{n^2} - 1}}{{{n^2}}}

Hence, the sum of the given expression is n2−1n2\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+41 + 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.