Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Let \[{S_n} = \dfrac{1}{{{1^3}}} + \dfrac{{1 + 2}}{{{1^3} + {2^3}}} + \dfrac{{1 + 2 + 3}}{{{1^3} + {2^3} + {3^3}}} + \ldots \ldots + \dfrac{{1 + 2 + \ldots + n}}{{{1^3} + {2^3} + \ldots + {n^3}}}\]. If \[100{S_n} = n\], then find the value of \[n\].
A). 199
B). 99
C). 200
D). 19

seo-qna
Last updated date: 22nd Jul 2024
Total views: 350.7k
Views today: 7.50k
Answer
VerifiedVerified
350.7k+ views
Hint: Here, in the question, we are given \[{S_n} = \dfrac{1}{{{1^3}}} + \dfrac{{1 + 2}}{{{1^3} + {2^3}}} + \dfrac{{1 + 2 + 3}}{{{1^3} + {2^3} + {3^3}}} + \ldots \ldots + \dfrac{{1 + 2 + \ldots + n}}{{{1^3} + {2^3} + \ldots + {n^3}}}\] which is the sum to \[n\] terms of a kind of special series. In order to find \[n\] such that \[100{S_n} = n\], we have to simplify \[{S_n}\] in terms of \[n\]. Then using the given equation \[100{S_n} = n\], we can find the value of \[n\].
Formulae used:
\[{S_n} = \sum\limits_{k = 1}^n {{a_k}} \], where \[{S_n}\] is the sum of \[n\] terms of the series and \[{a_k}\] is the \[{k^{th}}\] term of the series.
Sum of first \[n\] natural numbers=\[\dfrac{{n\left( {n + 1} \right)}}{2}\]
Sum of the cubes of first \[n\] natural numbers=\[{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\]

Complete step-by-step solution:
Let us collect the given information,
\[{S_n} = \dfrac{1}{{{1^3}}} + \dfrac{{1 + 2}}{{{1^3} + {2^3}}} + \dfrac{{1 + 2 + 3}}{{{1^3} + {2^3} + {3^3}}} + \ldots \ldots + \dfrac{{1 + 2 + \ldots + n}}{{{1^3} + {2^3} + \ldots + {n^3}}}\], and,
\[100{S_n} = n\]
Now, we have, \[{a_k} = \dfrac{{1 + 2 + 3 + \ldots + k}}{{{1^3} + {2^3} + {3^3} + \ldots + {k^3}}}\]
\[{S_n} = \dfrac{1}{{{1^3}}} + \dfrac{{1 + 2}}{{{1^3} + {2^3}}} + \dfrac{{1 + 2 + 3}}{{{1^3} + {2^3} + {3^3}}} + \ldots \ldots + \dfrac{{1 + 2 + \ldots + n}}{{{1^3} + {2^3} + \ldots + {n^3}}}\]
Observing carefully the series, we get the \[{k^{th}}\] term of the series as:
\[{a_k} = \dfrac{{1 + 2 + 3 + \ldots + k}}{{{1^3} + {2^3} + {3^3} + \ldots + {k^3}}}\]
Using identities,
Sum of first \[n\] natural numbers =\[\dfrac{{n\left( {n + 1} \right)}}{2}\]
Sum of the cubes of first \[n\] natural numbers =\[{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\], we get,
\[{a_k} = \dfrac{{\dfrac{{k\left( {k + 1} \right)}}{2}}}{{{{\left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right)}^2}}}\]
Simplifying it, we get,
\[ \Rightarrow {a_k} = \dfrac{2}{{k\left( {k + 1} \right)}}\]
\[ \Rightarrow {a_k} = \dfrac{2}{k} - \dfrac{2}{{k + 1}}\]
 Now, we have to find the Sum of \[n\] terms, which is \[{S_n}\] from the \[{k^{th}}\] term,
 Using the formula \[{S_n} = \sum\limits_{k = 1}^n {{a_k}} \], we get
\[{S_n} = \sum\limits_{k = 1}^n {\left( {\dfrac{2}{k} - \dfrac{2}{{k + 1}}} \right)} \]
Expanding this, we get
\[{S_n} = 2-\dfrac{2}{2}+1- \dfrac{2}{3} + \dfrac{2}{3} - \dfrac{2}{4} + \cdots + \dfrac{2}{{n - 1}} - \dfrac{2}{n} + \dfrac{2}{n} - \dfrac{2}{{n + 1}}\]
In the above expression we have an additive inverse of each term present except for the first and the last term. Therefore,
\[{S_n} = 2 - \dfrac{2}{{n + 1}} \\
   \Rightarrow {S_n} = \dfrac{{2n}}{{n + 1}} \]
Given that \[100{S_n} = n\]
Putting the value of \[{S_n}\], we get
\[ 100 \times \dfrac{{2n}}{{n + 1}} = n \\
   \Rightarrow 200n = n\left( {n + 1} \right) \\
   \Rightarrow n + 1 = 200 \\
   \Rightarrow n = 199 \]
Hence, the value of \[n\] is \[199\].
Hence option A. \[199\] is the correct answer.

Note: The series given in the question is a special kind of series. Special series are the series which are special in some or other way. It might be arithmetic or geometric or any other type of progressive series. While solving such types of questions, we must find \[{k^{th}}\] otherwise it would be very hectic to solve. With the help of \[{k^{th}}\] term, we can find \[{S_n}\] easily.