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# 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 $2126f\left( {1063} \right)$.

Last updated date: 11th Aug 2024
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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 $2126f\left( {1063} \right)$.

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 $f\left( 1 \right) = 1$.