Question

The sum of the series $\sum\limits_{n=8}^{17}{\dfrac{1}{(n+2)(n+3)}}$ is equal to
A. $\dfrac{1}{17}$
B. $\dfrac{1}{18}$
C. $\dfrac{1}{19}$
D. $\dfrac{1}{20}$

Answer 1

Hint : In these types of questions, firstly we have to check which value we can substitute to make it simpler for us, and after simplifying it we have to substitute the given values of $n$. After substituting all the values we have to simplify by performing required arithmetic operations and we will get our required result.

Complete step-by-step solution:
A series can simply be defined as the sum of the various numbers, or elements of a sequence. The series can be finite or infinite depending on the sequence whether it is finite or infinite.
Sequence can be defined as the set of the elements that follow a certain pattern whereas series can be defined as the sum of elements of the given sequence. The finite series are series where the numbers are ending and infinite series are the series where the numbers are never ending.
Types of series are as follows:
Geometric Series
Harmonic Series
Power Series
Alternating Series
Exponent Series
A geometric series can be defined as series with a constant ratio between successive terms.
A harmonic series can be defined as the series that contains the sum of terms that are the reciprocal of the arithmetic series terms.
Power series can be defined as the series that can be thought of as a polynomial with an infinite number of terms.
As, we have given in the question $\sum\limits_{n=8}^{17}{\dfrac{1}{(n+2)(n+3)}}$
So, firstly in this we will substitute $1$ with $(n+3)-(n+2)$
Now, it becomes:
$\sum\limits_{n=8}^{17}{\dfrac{(n+3)-(n+2)}{(n+2)(n+3)}}$
$\Rightarrow \sum\limits_{n=8}^{17}{\dfrac{(n+3)}{(n+2)(n+3)}-\dfrac{(n+2)}{(n+2)(n+3)}}$
After cancelling the same part in numerator and denominator it becomes as:
$\sum\limits_{n=8}^{17}{\dfrac{1}{(n+2)}-\dfrac{1}{(n+3)}}$
$\Rightarrow \sum\limits_{n=8}^{17}{\dfrac{1}{(n+2)}-\sum\limits_{n=8}^{17}{\dfrac{1}{(n+3)}}}$
Now we will substitute $n$ from $8$ to $17$
After substituting the values, it becomes:
$=[\dfrac{1}{10}-\dfrac{1}{11}]+[\dfrac{1}{11}-\dfrac{1}{12}]+............+[\dfrac{1}{18}-\dfrac{1}{19}]+[\dfrac{1}{19}-\dfrac{1}{20}]$
Now we will solve this:
We will cancel the terms that have same value but have different signs such as positive and negative.
$=\dfrac{1}{10}-\dfrac{1}{20}$
Taking LCM of both the terms and then subtracting both the terms:

& =\dfrac{2-1}{20} \\\ & =\dfrac{1}{20} \\\ \end{aligned}$$ So, we can say that the correct option is $$4$$. **Note:** There is a series called as Fibonacci series in which when we add the last two numbers, then the sum of last two numbers is treated as the next number in the series ( i.e., each number is obtained by adding the two preceding numbers. Using the Pascal’s triangle, Fibonacci numbers can be obtained.