Let \[f\] be a function from positive integers to the set of real numbers such that
(i) \[f\left( 1 \right) = 1\]
(ii) \[\sum\limits_{r = 1}^n {rf\left( r \right)} = n\left( {n + 1} \right)f\left( n \right),{\rm{ }}\forall n \ge 2\]
Find the value of \[2126f\left( {1063} \right)\].

Hint: Here, we will use the given information to find the values of the function at 2, 3, 4, and 5. Then, we will use these equations to form a generalised equation. Finally, we will use the generalised equation to find and simplify the value of the expression \[2126f\left( {1063} \right)\].

Complete step-by-step answer:
It is given that \[n \ge 2\].
Substituting \[n = 2\] in the equation \[\sum\limits_{r = 1}^n {rf\left( r \right)} = n\left( {n + 1} \right)f\left( n \right)\], we get
\[ \Rightarrow \sum\limits_{r = 1}^2 {rf\left( r \right)} = 2\left( {2 + 1} \right)f\left( 2 \right)\]
Adding the terms in the expression, we get
\[ \Rightarrow \sum\limits_{r = 1}^2 {rf\left( r \right)} = 2\left( 3 \right)f\left( 2 \right)\]
Multiplying the terms in the expression, we get
\[ \Rightarrow \sum\limits_{r = 1}^2 {rf\left( r \right)} = 6f\left( 2 \right)\]
Expanding the sum, we get
\[ \Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) = 6f\left( 2 \right)\]
Subtracting \[2f\left( 2 \right)\] from both sides, we get
\[\begin{array}{l} \Rightarrow f\left( 1 \right) + 2f\left( 2 \right) - 2f\left( 2 \right) = 6f\left( 2 \right) - 2f\left( 2 \right)\\ \Rightarrow f\left( 1 \right) = 4f\left( 2 \right)\end{array}\]
Substituting \[f\left( 1 \right) = 1\] in the equation, we get
\[ \Rightarrow 1 = 4f\left( 2 \right)\]
Dividing both sides of the equation by 4, we get
\[ \Rightarrow f\left( 2 \right) = \dfrac{1}{4} \ldots \ldots \ldots \left( 1 \right)\]
Substituting \[n = 3\] in the equation \[\sum\limits_{r = 1}^n {rf\left( r \right)} = n\left( {n + 1} \right)f\left( n \right)\], we get
\[ \Rightarrow \sum\limits_{r = 1}^3 {rf\left( r \right)} = 3\left( {3 + 1} \right)f\left( 3 \right)\]
Adding the terms in the expression, we get
\[ \Rightarrow \sum\limits_{r = 1}^3 {rf\left( r \right)} = 3\left( 4 \right)f\left( 3 \right)\]
Multiplying the terms in the expression, we get
\[ \Rightarrow \sum\limits_{r = 1}^3 {rf\left( r \right)} = 12f\left( 3 \right)\]
Expanding the sum, we get
\[ \Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) + 3f\left( 3 \right) = 12f\left( 3 \right)\]
Subtracting \[3f\left( 3 \right)\] from both sides, we get
\[\begin{array}{l} \Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) + 3f\left( 3 \right) - 3f\left( 3 \right) = 12f\left( 3 \right) - 3f\left( 3 \right)\\ \Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) = 9f\left( 3 \right)\end{array}\]
Substituting \[f\left( 1 \right) = 1\] and \[f\left( 2 \right) = \dfrac{1}{4}\] in the equation, we get
\[ \Rightarrow 1\left( 1 \right) + 2\left( {\dfrac{1}{4}} \right) = 9f\left( 3 \right)\]
Multiplying and adding the terms of the expression, we get
\[\begin{array}{l} \Rightarrow 1 + \dfrac{1}{2} = 9f\left( 3 \right)\\ \Rightarrow \dfrac{3}{2} = 9f\left( 3 \right)\end{array}\]
Dividing both sides by 6, we get
\[\begin{array}{l} \Rightarrow f\left( 3 \right) = \dfrac{3}{{2 \times 9}}\\ \Rightarrow f\left( 3 \right) = \dfrac{1}{{2 \times 3}}\\ \Rightarrow f\left( 3 \right) = \dfrac{1}{6} \ldots \ldots \ldots \left( 2 \right)\end{array}\]
Substituting \[n = 4\] in the equation \[\sum\limits_{r = 1}^n {rf\left( r \right)} = n\left( {n + 1} \right)f\left( n \right)\], we get
\[ \Rightarrow \sum\limits_{r = 1}^4 {rf\left( r \right)} = 4\left( {4 + 1} \right)f\left( 4 \right)\]
Adding the terms in the expression, we get
\[ \Rightarrow \sum\limits_{r = 1}^4 {rf\left( r \right)} = 4\left( 5 \right)f\left( 4 \right)\]
Multiplying the terms in the expression, we get
\[ \Rightarrow \sum\limits_{r = 1}^4 {rf\left( r \right)} = 20f\left( 4 \right)\]
Expanding the sum, we get
\[ \Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) + 3f\left( 3 \right) + 4f\left( 4 \right) = 20f\left( 4 \right)\]
Subtracting \[4f\left( 4 \right)\] from both sides, we get
\[\begin{array}{l} \Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) + 3f\left( 3 \right) + 4f\left( 4 \right) - 4f\left( 4 \right) = 20f\left( 4 \right) - 4f\left( 4 \right)\\ \Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) + 3f\left( 3 \right) = 16f\left( 4 \right)\end{array}\]
Substituting \[f\left( 1 \right) = 1\], \[f\left( 2 \right) = \dfrac{1}{4}\], and \[f\left( 3 \right) = \dfrac{1}{6}\] in the equation, we get
\[ \Rightarrow 1\left( 1 \right) + 2\left( {\dfrac{1}{4}} \right) + 3\left( {\dfrac{1}{6}} \right) = 16f\left( 4 \right)\]
Multiplying and adding the terms of the expression, we get
\[\begin{array}{l} \Rightarrow 1 + \dfrac{1}{2} + \dfrac{1}{2} = 16f\left( 4 \right)\\ \Rightarrow 2 = 16f\left( 4 \right)\end{array}\]
Dividing both sides by 16, we get
\[\begin{array}{l} \Rightarrow f\left( 4 \right) = \dfrac{2}{{16}}\\ \Rightarrow f\left( 4 \right) = \dfrac{1}{8} \ldots \ldots \ldots \left( 3 \right)\end{array}\]
Substituting \[n = 5\] in the equation \[\sum\limits_{r = 1}^n {rf\left( r \right)} = n\left( {n + 1} \right)f\left( n \right)\], we get
\[ \Rightarrow \sum\limits_{r = 1}^5 {rf\left( r \right)} = 5\left( {5 + 1} \right)f\left( 5 \right)\]
Adding the terms in the expression, we get
\[ \Rightarrow \sum\limits_{r = 1}^5 {rf\left( r \right)} = 5\left( 6 \right)f\left( 5 \right)\]
Multiplying the terms in the expression, we get
\[ \Rightarrow \sum\limits_{r = 1}^5 {rf\left( r \right)} = 30f\left( 5 \right)\]
Expanding the sum, we get
\[ \Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) + 3f\left( 3 \right) + 4f\left( 4 \right) + 5f\left( 5 \right) = 30f\left( 5 \right)\]
Subtracting \[5f\left( 5 \right)\] from both sides, we get
\[\begin{array}{l} \Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) + 3f\left( 3 \right) + 4f\left( 4 \right) + 5f\left( 5 \right) - 5f\left( 5 \right) = 30f\left( 5 \right) - 5f\left( 5 \right)\\ \Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) + 3f\left( 3 \right) + 4f\left( 4 \right) = 25f\left( 5 \right)\end{array}\]
Substituting \[f\left( 1 \right) = 1\], \[f\left( 2 \right) = \dfrac{1}{4}\], \[f\left( 3 \right) = \dfrac{1}{6}\], and \[f\left( 4 \right) = \dfrac{1}{8}\] in the equation, we get
\[ \Rightarrow 1\left( 1 \right) + 2\left( {\dfrac{1}{4}} \right) + 3\left( {\dfrac{1}{6}} \right) + 4\left( {\dfrac{1}{8}} \right) = 25f\left( 5 \right)\]
Multiplying and adding the terms of the expression, we get
\[\begin{array}{l} \Rightarrow 1 + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} = 25f\left( 5 \right)\\ \Rightarrow \dfrac{5}{2} = 25f\left( 5 \right)\end{array}\]
Dividing both sides by 25, we get
\[\begin{array}{l} \Rightarrow f\left( 5 \right) = \dfrac{5}{{2 \times 25}}\\ \Rightarrow f\left( 5 \right) = \dfrac{1}{{2 \times 5}}\\ \Rightarrow f\left( 5 \right) = \dfrac{1}{{10}} \ldots \ldots \ldots \left( 4 \right)\end{array}\]
Now, from equations \[\left( 1 \right)\], \[\left( 2 \right)\], \[\left( 3 \right)\], and \[\left( 4 \right)\], we have
\[f\left( 2 \right) = \dfrac{1}{4}\]
\[f\left( 3 \right) = \dfrac{1}{6}\]
\[f\left( 4 \right) = \dfrac{1}{8}\]
\[f\left( 5 \right) = \dfrac{1}{{10}}\]
Rewriting these equations, we get
\[f\left( 2 \right) = \dfrac{1}{{2 \times 2}}\]
\[f\left( 3 \right) = \dfrac{1}{{2 \times 3}}\]
\[f\left( 4 \right) = \dfrac{1}{{2 \times 4}}\]
\[f\left( 5 \right) = \dfrac{1}{{2 \times 5}}\]
Thus, from the above equations, we can generalise the formula for \[f\left( n \right)\] as
\[f\left( n \right) = \dfrac{1}{{2n}}\]
Now, we will find the value of \[f\left( {1063} \right)\].
Substituting \[n = 1063\] in the generalised equation \[f\left( n \right) = \dfrac{1}{{2n}}\], we get
\[ \Rightarrow f\left( {1063} \right) = \dfrac{1}{{2 \times 1063}}\]
Multiplying the terms of the expression, we get
\[ \Rightarrow f\left( {1063} \right) = \dfrac{1}{{2126}}\]
Multiplying both sides by 2126, we get
\[ \Rightarrow 2126f\left( {1063} \right) = \dfrac{{2126}}{{2126}}\]
Simplifying the expression, we get
\[ \Rightarrow 2126f\left( {1063} \right) = 1\]
\[\therefore \] The value of the expression \[2126f\left( {1063} \right)\] is 1.

Note: We need to take care of the values that \[n\] can take according to the question. We substituted the values of \[n\] as 2, 3, 4, and 5, and used the equations to form the generalised equation. We did not substitute the value of \[n\] as 1 because it is given that \[n \ge 2\]. If we substitute \[n\] as 1 in \[\sum\limits_{r = 1}^n {rf\left( r \right)} = n\left( {n + 1} \right)f\left( n \right)\], then we get the equation \[f\left( 1 \right) = 2f\left( 1 \right)\]. This is not possible since \[f\left( 1 \right) = 1\].