Question

f(x) is a polynomial of degree 2020. f(k) = $\frac{2}{k}$ , where k = 1, 2, 3, ...., 2021. Find f(2022).

Answer 1

$\textbf{Given:}$

$f(k)=\dfrac{2}{k}$

$\textbf{To find:}$

$\text{The value of f(2022)}$

$\textbf{Solution:}$

$\text{Consider,}$

$f(k)=\dfrac{2}{k}$     ...................(1)

$\implies\,f(k+1)=\dfrac{2}{k+1}$      .............(2)

$\text{Divide (2) by (1), we get}$

$\dfrac{f(k+1)}{f(k)}=\dfrac{\dfrac{2}{k+1}}{\dfrac{2}{k}}$

$\dfrac{f(k+1)}{f(k)}=\dfrac{\dfrac{1}{k+1}}{\dfrac{1}{k}}$

$\dfrac{f(k+1)}{f(k)}=\dfrac{k}{k+1}$

$\implies\bf\,f(k+1)=f(k)(\dfrac{k}{k+1})$

$\text{Put k=2021}$

$f(2022)=f(2021)(\dfrac{2021}{2021+1})$

$f(2022)=\dfrac{2}{2021}(\dfrac{2021}{2022})$

$f(2022)=\dfrac{2}{2022}$

$\bf\,f(2022)=\dfrac{1}{1011}$

$\therefore\textbf{The value of f(2022) is \bf\dfrac{1}{1011} }$

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