Question

If we have a series $\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+......$to n terms $=\dfrac{1}{18}-f(n),$then find $f(n)$:
(A) $\dfrac{1}{3\left( n+2 \right)\left( n+3 \right)}$
(B) $\dfrac{1}{3\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}$
(C) $\dfrac{1}{6\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}$
(D) $\dfrac{1}{6\left( n+2 \right)\left( n+3 \right)}$

Answer 1

To solve this type of questions first write general term in factor form of this series. We have to Write that the general term in the form of $T=k\left( \dfrac{1}{first\left( n-1 \right)terms}-\dfrac{1}{last\left( n-1 \right)terms} \right)$. And then solve it. Firstly, we have to take the L.H.S. of the equation and then we will put the given information in this ${{S}_{n}}=\sum\limits_{r=1}^{n}{\left( \dfrac{1}{r\left( r+1 \right)\left( r+2 \right)\left( r+3 \right)} \right)}$ Formula and then we will compare it with the question and then we will get the value of the $f(n)$.

Complete step-by-step solution:
Let’s start our problem with the given data firstly we will take the L.H.S. of the equation Lets consider it as the ${{S}_{n}}$.
${{S}_{n}}=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+......$up to n terms
And we have the formula,
${{S}_{n}}=\sum\limits_{r=1}^{n}{{{T}_{r}}}$
We will Write ${{T}_{r}}$ in the general ${{r}^{th}}$term of the sum of this series we will get,
${{T}_{r}}=\dfrac{1}{r\left( r+1 \right)\left( r+2 \right)\left( r+3 \right)}$
After this combine all the equations in the form of ${{S}_{n}}$ and we will get the combine formula as,
${{S}_{n}}=\sum\limits_{r=1}^{n}{\left( \dfrac{1}{r\left( r+1 \right)\left( r+2 \right)\left( r+3 \right)} \right)}$
As,
${{T}_{r}}=\dfrac{1}{r\left( r+1 \right)\left( r+2 \right)\left( r+3 \right)}$
Then with the help of these formulas we will Separate the factors up to the n terms are given below,
${{T}_{r}}=\dfrac{1}{3}\left( \dfrac{1}{r\left( r+1 \right)\left( r+2 \right)}-\dfrac{1}{\left( r+1 \right)\left( r+2 \right)\left( r+3 \right)} \right)$

&\Rightarrow {{T}_{1}}=\dfrac{1}{3}\left( \dfrac{1}{1.2.3}-\dfrac{1}{2.3.4} \right) \\\ &\Rightarrow {{T}_{2}}=\dfrac{1}{3}\left( \dfrac{1}{2.3.4}-\dfrac{1}{3.4.5} \right) \\\ &\Rightarrow {{T}_{3}}=\dfrac{1}{3}\left( \dfrac{1}{3.4.5}-\dfrac{1}{4.5.6} \right) \\\ &\Rightarrow {{T}_{n-1}}=\dfrac{1}{3}\left( \dfrac{1}{\left( n-1 \right)\left( n \right)\left( n+1 \right)}-\dfrac{1}{\left( n \right)\left( n+1 \right)\left( n+2 \right)} \right) \\\ &\Rightarrow {{T}_{n}}=\dfrac{1}{3}\left( \dfrac{1}{n\left( n+1 \right)\left( n+2 \right)}-\dfrac{1}{\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)} \right) \\\ \end{aligned}$$ After combining all these terms Then we will get $${{S}_{n}}$$ as, $${{S}_{n}}=\sum\limits_{r=1}^{n}{{{T}_{r}}}$$ $$\Rightarrow {{S}_{n}}=\dfrac{1}{3}\left( \dfrac{1}{1.2.3}-\dfrac{1}{\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)} \right)$$ But in the question, we have Given that $${{S}_{n}}=\dfrac{1}{18}-f\left( n \right)$$ so now let’s equate it. $${{S}_{n}}=\dfrac{1}{18}-f\left( n \right)=\dfrac{1}{3}\left( \dfrac{1}{1.2.3}-\dfrac{1}{\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)} \right)$$ After equating we will get the equation as, $$\dfrac{1}{18}-f\left( n \right)=\left( \dfrac{1}{18}-\dfrac{1}{3\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)} \right)$$ After solving the above equation, we will get, $$f\left( n \right)=\dfrac{1}{3\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}$$ So, the above equation matches with the option B) **Hence the Option (B) is the correct option.** **Note:** The above method is only applicable when we have the question which is making the sequence and when you are doing the factors don’t forget to take the k as the constant. In these types of problems mostly the students are getting confused by looking at the question on the first site. We have to solve this problem by considering the RHS and LHS of the equation by this method most of the time we will get the answer.