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Question: Let \[f\] be a function from positive integers to the set of real numbers such that (i) \[f\left( ...

Let ff be a function from positive integers to the set of real numbers such that
(i) f(1)=1f\left( 1 \right) = 1
(ii) ∑r=1nrf(r)=n(n+1)f(n),∀n≥2\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(1063)2126f\left( {1063} \right).

Explanation

Solution

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

Complete step-by-step answer:
It is given that n≥2n \ge 2.
Substituting n=2n = 2 in the equation ∑r=1nrf(r)=n(n+1)f(n)\sum\limits_{r = 1}^n {rf\left( r \right)} = n\left( {n + 1} \right)f\left( n \right), we get
⇒∑r=12rf(r)=2(2+1)f(2)\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
⇒∑r=12rf(r)=2(3)f(2)\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
⇒∑r=12rf(r)=6f(2)\Rightarrow \sum\limits_{r = 1}^2 {rf\left( r \right)} = 6f\left( 2 \right)
Expanding the sum, we get
⇒1f(1)+2f(2)=6f(2)\Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) = 6f\left( 2 \right)
Subtracting 2f(2)2f\left( 2 \right) from both sides, we get
⇒f(1)+2f(2)−2f(2)=6f(2)−2f(2) ⇒f(1)=4f(2)\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(1)=1f\left( 1 \right) = 1 in the equation, we get
⇒1=4f(2)\Rightarrow 1 = 4f\left( 2 \right)
Dividing both sides of the equation by 4, we get
⇒f(2)=14………(1)\Rightarrow f\left( 2 \right) = \dfrac{1}{4} \ldots \ldots \ldots \left( 1 \right)
Substituting n=3n = 3 in the equation ∑r=1nrf(r)=n(n+1)f(n)\sum\limits_{r = 1}^n {rf\left( r \right)} = n\left( {n + 1} \right)f\left( n \right), we get
⇒∑r=13rf(r)=3(3+1)f(3)\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
⇒∑r=13rf(r)=3(4)f(3)\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
⇒∑r=13rf(r)=12f(3)\Rightarrow \sum\limits_{r = 1}^3 {rf\left( r \right)} = 12f\left( 3 \right)
Expanding the sum, we get
⇒1f(1)+2f(2)+3f(3)=12f(3)\Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) + 3f\left( 3 \right) = 12f\left( 3 \right)
Subtracting 3f(3)3f\left( 3 \right) from both sides, we get
⇒1f(1)+2f(2)+3f(3)−3f(3)=12f(3)−3f(3) ⇒1f(1)+2f(2)=9f(3)\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(1)=1f\left( 1 \right) = 1 and f(2)=14f\left( 2 \right) = \dfrac{1}{4} in the equation, we get
⇒1(1)+2(14)=9f(3)\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
⇒1+12=9f(3) ⇒32=9f(3)\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
⇒f(3)=32×9 ⇒f(3)=12×3 ⇒f(3)=16………(2)\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=4n = 4 in the equation ∑r=1nrf(r)=n(n+1)f(n)\sum\limits_{r = 1}^n {rf\left( r \right)} = n\left( {n + 1} \right)f\left( n \right), we get
⇒∑r=14rf(r)=4(4+1)f(4)\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
⇒∑r=14rf(r)=4(5)f(4)\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
⇒∑r=14rf(r)=20f(4)\Rightarrow \sum\limits_{r = 1}^4 {rf\left( r \right)} = 20f\left( 4 \right)
Expanding the sum, we get
⇒1f(1)+2f(2)+3f(3)+4f(4)=20f(4)\Rightarrow 1f\left( 1 \right) + 2f\left( 2 \right) + 3f\left( 3 \right) + 4f\left( 4 \right) = 20f\left( 4 \right)
Subtracting 4f(4)4f\left( 4 \right) from both sides, we get
⇒1f(1)+2f(2)+3f(3)+4f(4)−4f(4)=20f(4)−4f(4) ⇒1f(1)+2f(2)+3f(3)=16f(4)\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(1)=1f\left( 1 \right) = 1, f(2)=14f\left( 2 \right) = \dfrac{1}{4}, and f(3)=16f\left( 3 \right) = \dfrac{1}{6} in the equation, we get
⇒1(1)+2(14)+3(16)=16f(4)\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
⇒1+12+12=16f(4) ⇒2=16f(4)\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
⇒f(4)=216 ⇒f(4)=18………(3)\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=5n = 5 in the equation ∑r=1nrf(r)=n(n+1)f(n)\sum\limits_{r = 1}^n {rf\left( r \right)} = n\left( {n + 1} \right)f\left( n \right), we get
⇒∑r=15rf(r)=5(5+1)f(5)\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
⇒∑r=15rf(r)=5(6)f(5)\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
⇒∑r=15rf(r)=30f(5)\Rightarrow \sum\limits_{r = 1}^5 {rf\left( r \right)} = 30f\left( 5 \right)
Expanding the sum, we get
⇒1f(1)+2f(2)+3f(3)+4f(4)+5f(5)=30f(5)\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(5)5f\left( 5 \right) from both sides, we get
⇒1f(1)+2f(2)+3f(3)+4f(4)+5f(5)−5f(5)=30f(5)−5f(5) ⇒1f(1)+2f(2)+3f(3)+4f(4)=25f(5)\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(1)=1f\left( 1 \right) = 1, f(2)=14f\left( 2 \right) = \dfrac{1}{4}, f(3)=16f\left( 3 \right) = \dfrac{1}{6}, and f(4)=18f\left( 4 \right) = \dfrac{1}{8} in the equation, we get
⇒1(1)+2(14)+3(16)+4(18)=25f(5)\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
⇒1+12+12+12=25f(5) ⇒52=25f(5)\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
⇒f(5)=52×25 ⇒f(5)=12×5 ⇒f(5)=110………(4)\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 (1)\left( 1 \right), (2)\left( 2 \right), (3)\left( 3 \right), and (4)\left( 4 \right), we have
f(2)=14f\left( 2 \right) = \dfrac{1}{4}
f(3)=16f\left( 3 \right) = \dfrac{1}{6}
f(4)=18f\left( 4 \right) = \dfrac{1}{8}
f(5)=110f\left( 5 \right) = \dfrac{1}{{10}}
Rewriting these equations, we get
f(2)=12×2f\left( 2 \right) = \dfrac{1}{{2 \times 2}}
f(3)=12×3f\left( 3 \right) = \dfrac{1}{{2 \times 3}}
f(4)=12×4f\left( 4 \right) = \dfrac{1}{{2 \times 4}}
f(5)=12×5f\left( 5 \right) = \dfrac{1}{{2 \times 5}}
Thus, from the above equations, we can generalise the formula for f(n)f\left( n \right) as
f(n)=12nf\left( n \right) = \dfrac{1}{{2n}}
Now, we will find the value of f(1063)f\left( {1063} \right).
Substituting n=1063n = 1063 in the generalised equation f(n)=12nf\left( n \right) = \dfrac{1}{{2n}}, we get
⇒f(1063)=12×1063\Rightarrow f\left( {1063} \right) = \dfrac{1}{{2 \times 1063}}
Multiplying the terms of the expression, we get
⇒f(1063)=12126\Rightarrow f\left( {1063} \right) = \dfrac{1}{{2126}}
Multiplying both sides by 2126, we get
⇒2126f(1063)=21262126\Rightarrow 2126f\left( {1063} \right) = \dfrac{{2126}}{{2126}}
Simplifying the expression, we get
⇒2126f(1063)=1\Rightarrow 2126f\left( {1063} \right) = 1
∴\therefore The value of the expression 2126f(1063)2126f\left( {1063} \right) is 1.

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