Answer
Verified
405k+ views
Hint: Here, the sum of $n$ consecutive terms of an expression is given as $\sum\limits_{r = 1}^n {{t_r}} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8}$. We have to firstly find the term ${t_r}$ by using formula ${t_r} = {S_n} - {S_{n - 1}}$. write $\dfrac{1}{{{t_r}}}$ and arrange them in suitable form then apply $\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} $ and we will get that the successive terms cancel each others.
Complete step-by-step solution:
Here, it is given that ${S_n} = \sum\limits_{r = 1}^n {{t_r}} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8}$ .
Now, we can find the term ${t_n}$ using the above given formula.
$
\Rightarrow {t_n} = {S_n} - {S_{n - 1}} \\
\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8} - \dfrac{{\left( {n - 1} \right)\left( {n + 1 - 1} \right)\left( {n + 2 - 1} \right)\left( {n + 3 - 1} \right)}}{8} \\
\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8} - \dfrac{{\left( {n - 1} \right)n\left( {n + 1} \right)\left( {n + 2} \right)}}{8}
$
Here, it is clearly visible that $\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{8}$ is present in both the terms so, we can take this as common and we can write,
$
\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{8}\left( {\left( {n + 3} \right) - \left( {n - 1} \right)} \right) \\
\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{2}
$
So, ${r^{th}}$ term of the required expression is ${t_r} = \dfrac{{r\left( {r + 1} \right)\left( {r + 2} \right)}}{2}$.
Now, we have to find the value of $\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} $. So, firstly write,
$ \Rightarrow \dfrac{1}{{{t_r}}} = \dfrac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}$
We have to add and subtract $2$ in the numerator of the term $\dfrac{1}{{{t_r}}}$ so that this can be converted into suitable formats.
$ \Rightarrow \dfrac{1}{{{t_r}}} = \dfrac{{\left( {r + 2} \right) - r}}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}$
We can write this as the difference of two fractions. That is
$ \Rightarrow \dfrac{1}{{{t_r}}} = \dfrac{1}{{r\left( {r + 1} \right)}} - \dfrac{1}{{\left( {r + 1} \right)\left( {r + 2} \right)}}$
So, $\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} = \dfrac{1}{{r\left( {r + 1} \right)}} - \dfrac{1}{{\left( {r + 1} \right)\left( {r + 2} \right)}}$
Here, we have to find the summation of $n$ terms, so we have to put the value of $r$ from $1$ to $n$.
$\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} = \dfrac{1}{{1\left( {1 + 1} \right)}} - \dfrac{1}{{\left( {1 + 1} \right)\left( {1 + 2} \right)}} + \dfrac{1}{{2 \times 3}} - \dfrac{1}{{3 \times 4}} - - - - - - - - + \dfrac{1}{{n\left( {n + 1} \right)}} - \dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$
It is clearly visible that ${2^{nd}}$ and ${3^{rd}}$ terms are cancelling each other and similarly next two terms cancel each other and finally only the first and last terms will be remaining. This imply
$ \Rightarrow \sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}} = \dfrac{1}{2} - \dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}} $
Taking $ - 1$ as common we can write,
$ \Rightarrow \sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} = - \left( {\dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}} - \dfrac{1}{2}} \right)$
Thus, option (A) is the correct answer.
Note: While solving the problem of summation of sequences and series we have to first write ${r^{th}}$ term and then convert this into suitable form so that except some terms others are cancelled out.
If the denominator of ${r^{th}}$term is cubic like \[\dfrac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}\] we can write
$ \Rightarrow \dfrac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}} = \dfrac{A}{n} + \dfrac{B}{{n + 1}} + \dfrac{C}{{n + 2}}$ and by equating on both side of equation we can get the value of $A,B$ and $C$ and then do summation as shown above.
Complete step-by-step solution:
Here, it is given that ${S_n} = \sum\limits_{r = 1}^n {{t_r}} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8}$ .
Now, we can find the term ${t_n}$ using the above given formula.
$
\Rightarrow {t_n} = {S_n} - {S_{n - 1}} \\
\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8} - \dfrac{{\left( {n - 1} \right)\left( {n + 1 - 1} \right)\left( {n + 2 - 1} \right)\left( {n + 3 - 1} \right)}}{8} \\
\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8} - \dfrac{{\left( {n - 1} \right)n\left( {n + 1} \right)\left( {n + 2} \right)}}{8}
$
Here, it is clearly visible that $\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{8}$ is present in both the terms so, we can take this as common and we can write,
$
\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{8}\left( {\left( {n + 3} \right) - \left( {n - 1} \right)} \right) \\
\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{2}
$
So, ${r^{th}}$ term of the required expression is ${t_r} = \dfrac{{r\left( {r + 1} \right)\left( {r + 2} \right)}}{2}$.
Now, we have to find the value of $\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} $. So, firstly write,
$ \Rightarrow \dfrac{1}{{{t_r}}} = \dfrac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}$
We have to add and subtract $2$ in the numerator of the term $\dfrac{1}{{{t_r}}}$ so that this can be converted into suitable formats.
$ \Rightarrow \dfrac{1}{{{t_r}}} = \dfrac{{\left( {r + 2} \right) - r}}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}$
We can write this as the difference of two fractions. That is
$ \Rightarrow \dfrac{1}{{{t_r}}} = \dfrac{1}{{r\left( {r + 1} \right)}} - \dfrac{1}{{\left( {r + 1} \right)\left( {r + 2} \right)}}$
So, $\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} = \dfrac{1}{{r\left( {r + 1} \right)}} - \dfrac{1}{{\left( {r + 1} \right)\left( {r + 2} \right)}}$
Here, we have to find the summation of $n$ terms, so we have to put the value of $r$ from $1$ to $n$.
$\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} = \dfrac{1}{{1\left( {1 + 1} \right)}} - \dfrac{1}{{\left( {1 + 1} \right)\left( {1 + 2} \right)}} + \dfrac{1}{{2 \times 3}} - \dfrac{1}{{3 \times 4}} - - - - - - - - + \dfrac{1}{{n\left( {n + 1} \right)}} - \dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$
It is clearly visible that ${2^{nd}}$ and ${3^{rd}}$ terms are cancelling each other and similarly next two terms cancel each other and finally only the first and last terms will be remaining. This imply
$ \Rightarrow \sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}} = \dfrac{1}{2} - \dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}} $
Taking $ - 1$ as common we can write,
$ \Rightarrow \sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} = - \left( {\dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}} - \dfrac{1}{2}} \right)$
Thus, option (A) is the correct answer.
Note: While solving the problem of summation of sequences and series we have to first write ${r^{th}}$ term and then convert this into suitable form so that except some terms others are cancelled out.
If the denominator of ${r^{th}}$term is cubic like \[\dfrac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}\] we can write
$ \Rightarrow \dfrac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}} = \dfrac{A}{n} + \dfrac{B}{{n + 1}} + \dfrac{C}{{n + 2}}$ and by equating on both side of equation we can get the value of $A,B$ and $C$ and then do summation as shown above.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE