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Question: The sum of \( \dfrac{{{1^2}}}{{1.3}} + \dfrac{{{2^2}}}{{3.5}} + \dfrac{{{3^2}}}{{5.7}} + .... + \dfr...

The sum of 121.3+223.5+325.7+....+n2(2n−1)(2n+1)\dfrac{{{1^2}}}{{1.3}} + \dfrac{{{2^2}}}{{3.5}} + \dfrac{{{3^2}}}{{5.7}} + .... + \dfrac{{{n^2}}}{{(2n - 1)(2n + 1)}} is:
A) (2n2+1)2n+1\dfrac{{(2{n^2} + 1)}}{{2n + 1}}
B) n(n+1)2(2n+1)\dfrac{{n(n + 1)}}{{2(2n + 1)}}
C) (2n2+1)4(2n+1)\dfrac{{(2{n^2} + 1)}}{{4(2n + 1)}}
D) None of these

Explanation

Solution

We have to find the sum of the following series, as we can judge the sequence of the numbers by the last number in the sequence which is in terms of nn, and is the nth{n^{th}} term in the sequence. Since the denominator is factorized we will try to do the partial decomposition method to break this into partial fractions which can then be solved using simple methods for the sum upto nn terms.

Complete step by step answer:
We can judge from the sequence :
121.3+223.5+325.7+....+n2(2n−1)(2n+1)  \dfrac{{{1^2}}}{{1.3}} + \dfrac{{{2^2}}}{{3.5}} + \dfrac{{{3^2}}}{{5.7}} + .... + \dfrac{{{n^2}}}{{(2n - 1)(2n + 1)}} \\\ \\\
That the nth{n^{th}} term in the sequence is given by,
tn  = n2(2n−1)(2n+1){t_n}\; = {\text{ }}\dfrac{{{n^2}}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}
We will now break this by partial decomposition method as below:
tn  = 14 + 18(2n−1) −18(2n+1){t_{n\;}} = {\text{ }}\dfrac{1}{4}{\text{ }} + {\text{ }}\dfrac{1}{{8\left( {2n - 1} \right)}}{\text{ }} - \dfrac{1}{{8\left( {2n + 1} \right)}}
The sum will now depend on three factors first being the number 14\dfrac{1}{4} this will be simple as it will be added nn times.
The other two factors will be solved for first 22 or 33 iterations and we will see a patter emerge and then solve the question further.
Sn  = (n4) + (18)(1− 13 + 13 −15 + 15 −17 +….1(2n−1) − 1(2n+1)]{S_n}\; = {\text{ }}\left( {\dfrac{n}{4}} \right){\text{ }} + {\text{ }}\left( {\dfrac{1}{8}} \right)\left( {1 - {\text{ }}\dfrac{1}{3}{\text{ }} + {\text{ }}\dfrac{1}{3}{\text{ }} - \dfrac{1}{5}{\text{ }} + {\text{ }}\dfrac{1}{5}{\text{ }} - \dfrac{1}{7}{\text{ }} + \ldots .\dfrac{1}{{\left( {2n - 1} \right)}}{\text{ }}-{\text{ }}\dfrac{1}{{\left( {2n + 1} \right)}}} \right]
We thus see that on adding in the second term only the last expression remains all other are cancelled except 11 . thus the next step will be:
Sn=n4+ (18) [ 1 − 1(2n+1)]{S_n} = \dfrac{n}{4} + {\text{ }}\left( {\dfrac{1}{8}} \right){\text{ }}\left[ {{\text{ }}1{\text{ }}-{\text{ }}\dfrac{1}{{\left( {2n + 1} \right)}}} \right]
⇒Sn=n4+ (18) [ 2n+1−1(2n+1)]\Rightarrow {S_n} = \dfrac{n}{4} + {\text{ }}\left( {\dfrac{1}{8}} \right){\text{ }}\left[ {{\text{ }}\dfrac{{2n + 1 - 1}}{{\left( {2n + 1} \right)}}} \right]
⇒Sn=n4+ (18) [ 2n(2n+1)]\Rightarrow {S_n} = \dfrac{n}{4} + {\text{ }}\left( {\dfrac{1}{8}} \right){\text{ }}\left[ {{\text{ }}\dfrac{{2n}}{{\left( {2n + 1} \right)}}} \right]
Solving it further we get,
Sn=n4+ (14) [ n(2n+1)]{S_n} = \dfrac{n}{4} + {\text{ }}\left( {\dfrac{1}{4}} \right){\text{ }}\left[ {{\text{ }}\dfrac{n}{{\left( {2n + 1} \right)}}} \right]
⇒Sn=(14) [ n(2n+1)+n]\Rightarrow {S_n} = \left( {\dfrac{1}{4}} \right){\text{ }}\left[ {{\text{ }}\dfrac{n}{{\left( {2n + 1} \right)}} + n} \right]
⇒Sn=(14) [ n(2n+1)+n(2n+1)]\Rightarrow {S_n} = \left( {\dfrac{1}{4}} \right){\text{ }}\left[ {{\text{ }}\dfrac{{n(2n + 1) + n}}{{\left( {2n + 1} \right)}}} \right]
Solving the brackets we will get a quadratic equation as,
Sn=(14) [ 2n2+2n(2n+1)]{S_n} = \left( {\dfrac{1}{4}} \right){\text{ }}\left[ {{\text{ }}\dfrac{{2{n^2} + 2n}}{{\left( {2n + 1} \right)}}} \right]
⇒Sn=(12) [ n(n+1)(2n+1)]\Rightarrow {S_n} = \left( {\dfrac{1}{2}} \right){\text{ }}\left[ {{\text{ }}\dfrac{{n(n + 1)}}{{\left( {2n + 1} \right)}}} \right]
⇒Sn=n(n+1)2(2n+1)\Rightarrow {S_n} = \dfrac{{n(n + 1)}}{{2(2n + 1)}}
Hence, option (B) is the correct option.

Note:
We have done the partial decomposition of n2(2n−1)(2n+1)\dfrac{{{n^2}}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} as follows:
First the long division method render the fraction as,
n2(2n−1)(2n+1)=14+14(2n−1)(2n+1)\dfrac{{{n^2}}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \dfrac{1}{4} + \dfrac{{\dfrac{1}{4}}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}
Solving the second expression we get,
14(2n−1)(2n+1)=14(2n−1)(2n+1)=116n2−4\dfrac{{\dfrac{1}{4}}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \dfrac{1}{{4\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \dfrac{1}{{16{n^2} - 4}}
Factorizing the denominator we get,
116n2−4=14(2n−1)(2n+1)\dfrac{1}{{16{n^2} - 4}} = \dfrac{1}{{4\left( {2n - 1} \right)\left( {2n + 1} \right)}}
The partial fraction will now be done as:
14(2n−1)(2n+1)=A2n+1+B2n−1\dfrac{{\dfrac{1}{4}}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \dfrac{A}{{2n + 1}} + \dfrac{B}{{2n - 1}}
Solving we will get value of A and B as
A=−18A = - \dfrac{1}{8} and B=18B = \dfrac{1}{8}
Therefore we can write,
14(2n−1)(2n+1)=−182n+1+182n−1\dfrac{{\dfrac{1}{4}}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \dfrac{{ - \dfrac{1}{8}}}{{2n + 1}} + \dfrac{{\dfrac{1}{8}}}{{2n - 1}}
Since this was the second expression we were evaluating the full expression becomes:
n2(2n−1)(2n+1)=14+−182n+1+182n−1\dfrac{{{n^2}}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \dfrac{1}{4} + \dfrac{{ - \dfrac{1}{8}}}{{2n + 1}} + \dfrac{{\dfrac{1}{8}}}{{2n - 1}}