Question

If \[\sum\limits_1^5 {\dfrac{1}{{n(n + 1)(n + 2)(n + 3)}} = \dfrac{k}{3}} \], the k is equal to
A) \[\dfrac{{55}}{{336}}\]
B) \[\dfrac{{17}}{{105}}\]
C) \[\dfrac{{19}}{{112}}\]
D) \[\dfrac{1}{6}\]

Answer 1

Hint: Break the term mentioned inside the summation so that we can simplify it if possible and then if once it will get simplified then expand the summation as keep on putting the value of n from \[1\] to \[5\]. Hence, after completing all the summation carefully and simplifying it compares it with the R.H.S mentioned as \[\dfrac{k}{3}\] and finally the of k will be obtained.

Complete step by step solution: The given summation is as \[\sum\limits_1^5 {\dfrac{1}{{n(n + 1)(n + 2)(n + 3)}}} \]
So, now only considering \[\dfrac{1}{{n(n + 1)(n + 2)(n + 3)}}\]and splitting it in such a way that it can be represented as addition or subtraction of two different terms as,
First of all multiply the numerator an denominator by \[{\text{2}}\]
\[ = \dfrac{1}{2}(\dfrac{2}{{n(n + 1)(n + 2)(n + 3)}})\]
Now rearranging the numerator we get,
\[ = \dfrac{1}{2}\left( {\dfrac{{(n + 1)(n + 2) - (n + 3)(n)}}{{n(n + 1)(n + 2)(n + 3)}}} \right)\]
Separating the terms we get,
\[ = \dfrac{1}{2}\left( {\dfrac{1}{{n(n + 3)}} - \dfrac{1}{{(n + 1)(n + 2)}}} \right)\]
Now, further simplify both the parts of the above equation as
Let’s first solve, \[\dfrac{1}{{n(n + 3)}}\]
So, multiply and divide the above equation by \[{\text{3}}\],
\[ = (\dfrac{1}{3})\dfrac{3}{{n(n + 3)}}\]
On splitting the numerator in the terms of denominator, we get,
\[ = (\dfrac{1}{3})\dfrac{{(n + 3) - (n)}}{{n(n + 3)}}\]
On simplification we get,
\[
   = (\dfrac{1}{3})(\dfrac{1}{n} - \dfrac{1}{{n + 3}}) \\
   = \dfrac{1}{{3n}} - \dfrac{1}{{3n + 9}} \\
 \]
Now, simplify the second part of the above equation and break the numerator in the form of given denominators
\[
   = \dfrac{1}{{(n + 1)(n + 2)}} \\
   = \dfrac{{(n + 2) - (n + 1)}}{{(n + 1)(n + 2)}} \\
 \]
On simplification we get,
\[ = \dfrac{1}{{n + 1}} - \dfrac{1}{{n + 2}}\]
Hence, the final simplification will be as,
\[
  \dfrac{1}{2}(\dfrac{2}{{n(n + 1)(n + 2)(n + 3)}}) = \dfrac{1}{2}[(\dfrac{1}{{3n}} - \dfrac{1}{{3n + 9}}) - (\dfrac{1}{{n + 1}} - \dfrac{1}{{n + 2}})] \\
  \dfrac{1}{2}(\dfrac{2}{{n(n + 1)(n + 2)(n + 3)}}) = \dfrac{1}{2}[\dfrac{1}{{3n}} - \dfrac{1}{{3n + 9}} - \dfrac{1}{{n + 1}} + \dfrac{1}{{n + 2}}] \\
 \]
Hence, \[\dfrac{1}{{n(n + 1)(n + 2)(n + 3)}}\]can now be replaced with \[\dfrac{1}{2}[\dfrac{1}{{3n}} - \dfrac{1}{{3n + 9}} - \dfrac{1}{{n + 1}} + \dfrac{1}{{n + 2}}]\]
So now we require the value of
\[
  \sum\limits_1^5 {\dfrac{1}{2}[\dfrac{1}{{3n}} - \dfrac{1}{{3n + 9}} - \dfrac{1}{{n + 1}} + \dfrac{1}{{n + 2}}] = \dfrac{k}{3}} \\
   \Rightarrow \sum\limits_1^5 {[\dfrac{1}{{3n}} - \dfrac{1}{{3n + 9}} - \dfrac{1}{{n + 1}} + \dfrac{1}{{n + 2}}] = \dfrac{{2k}}{3}} \\
 \]
Hence, now put the value of n and go on factorizing the terms of the above equation as
\[ = \dfrac{1}{3}(1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5}) - \dfrac{1}{3}(\dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8}) - (\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6}) + (\dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7})\]
Hence on simplifying further to the above terms we can conclude that
\[ = \dfrac{1}{3}(1 + \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{6} - \dfrac{1}{7} - \dfrac{1}{8}) - (\dfrac{1}{2} - \dfrac{1}{7}) = \dfrac{{2k}}{3}\]
On further simplification we get,
\[ = \dfrac{1}{3}(\dfrac{{11}}{6} - \dfrac{1}{6} - \dfrac{{15}}{{56}}) - (\dfrac{5}{{14}}) = \dfrac{{2k}}{3}\]
\[ = \dfrac{1}{3}(\dfrac{5}{3} - \dfrac{{15}}{{56}}) - (\dfrac{5}{{14}}) = \dfrac{{2k}}{3}\]
Hence, take the L.C.M further and solving the above equation we can say that,
\[
   = \dfrac{1}{3}(\dfrac{{280 - 45}}{{3 \times 56}}) - (\dfrac{5}{{14}}) = \dfrac{{2k}}{3} \\
   = \dfrac{1}{3}(\dfrac{{235}}{{3 \times 56}}) - (\dfrac{5}{{14}}) = \dfrac{{2k}}{3} \\
 \]
Now, again take L.C.M and simplify the above equation,
\[
   \Rightarrow 2k = (\dfrac{{235}}{{3 \times 14 \times 4}}) - (\dfrac{{15}}{{14}}) \\
   \Rightarrow 2k = (\dfrac{{235}}{{3 \times 14 \times 4}}) - (\dfrac{{15 \times 12}}{{14 \times 12}}) \\
 \]
On simplification we get,
\[
   \Rightarrow 2k = \dfrac{{55}}{{168}} \\
   \Rightarrow k = \dfrac{{55}}{{336}} \\
 \]

Hence, we can see that as per the value of k option (a) is the correct answer.

Note: In mathematics, the summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total.
We can also break the given equation inside the summation through partial fraction method and can split the terms. In higher mathematics these question can be solved by some more advance concepts.