Question

Let $ f:\mathsf{\mathbb{N}}\to \mathsf{\mathbb{R}} $ be such that $ f\left( 1 \right)+2f\left( 2 \right)+3f\left( 3 \right)+...nf\left( n \right)=n\left( n+1 \right)f\left( n \right) $ for all $ n\in \mathsf{\mathbb{N}},n\ge 2 $ and $ f\left( 1 \right)=1 $ . If the value of $ f\left( 1003 \right)=\dfrac{1}{k} $ then $ k= $ \[\]
A. $ 1003 $ \[\]
B. $ 2003 $ \[\]
C. $ 2005 $ \[\]
D. $ 2006 $ \[\]

Answer 1

Hint: We recall from algebra of functions that if $ C $ is constant and $ f\left( x \right) $ is function then $ Cf\left( x \right) $ will have the same domain as $ f\left( x \right) $ . We choose a counter $ k=1,2,3,...n $ and put $ k=n-1 $ in the given condition then subtract respective sides to get $ kf\left( k \right) $ is constant for all $ k=1,2,3,...n $ . We use the given condition $ f\left( 1 \right)=1 $ and find a pattern by evaluating find $ f\left( 2 \right),f\left( 3 \right),...,f\left( n \right) $ . We find $ f\left( n \right) $ and then put $ n=1003 $ to find the answer. \[\]

Complete step by step answer:
We are given a function from natural number set $ \mathsf{\mathbb{N}} $ to the complex number set $ \mathsf{\mathbb{R}} $ which satisfies two conditions : $ f\left( 1 \right)=1 $ and
 $ f\left( 1 \right)+2f\left( 2 \right)+3f\left( 3 \right)+...nf\left( n \right)=n\left( n+1 \right)f\left( n \right).......\left( 1 \right) $
We assume a counter $ k $ which runs on integer’s $ 1,2,3...,n $ .We put $ k=n-1 $ and have the statement as;
 $ f\left( 1 \right)+2f\left( 2 \right)+3f\left( 3 \right)+...+\left( n-1 \right)f\left( n-1 \right)=n\left( n-1 \right)f\left( n-1 \right).......\left( 2 \right) $
We subtract respective sides of equation (2) from equation (1) to have;
\[nf\left( n \right)=n\left( n+1 \right)f\left( n \right)-n\left( n-1 \right)f\left( n-1 \right)\]
We divide all of the terms by $ n $ to have;
\[\begin{align}
  & f\left( n \right)=\left( n+1 \right)f\left( n \right)-\left( n-1 \right)f\left( n-1 \right) \\
 & \Rightarrow \left( n+1-1 \right)f\left( n \right)=\left( n-1 \right)f\left( n-1 \right) \\
 & \Rightarrow nf\left( n \right)=\left( n-1 \right)f\left( n-1 \right)........\left( 3 \right) \\
\end{align}\]
We see that $ kf\left( k \right) $ is a constant for all $ k=1,2,3,....n $ . We are given the values $ f\left( 1 \right)=1 $ in the question. Let us put $ n=2 $ since $ n $ is valid for $ n\ge 2 $ in above equation to have;
\[\begin{align}
  & 2\cdot f\left( 2 \right)=\left( 2-1 \right)f\left( 2-1 \right) \\
 & \Rightarrow 2f\left( 2 \right)=1\cdot f\left( 1 \right)=1\cdot 1 \\
 & \Rightarrow f\left( 2 \right)=\dfrac{1}{2} \\
\end{align}\]
We put $ n=3 $ in equation (3) to have;
\[\begin{align}
  & 3\cdot f\left( 3 \right)=\left( 3-1 \right)f\left( 3-1 \right) \\
 & \Rightarrow 3f\left( 3 \right)=3\cdot f\left( 2 \right)=2\times \dfrac{1}{2} \\
 & \Rightarrow f\left( 3 \right)=\dfrac{1}{3} \\
\end{align}\]
We observe that the function $ f\left( k \right) $ for all $ k=2,3...,n $ can be defined as;
\[f\left( k \right)=\dfrac{1}{k}\]
We are given that $ f\left( 1003 \right)=\dfrac{1}{k} $ . So we have;
\[f\left( 1003 \right)=\dfrac{1}{1003}\]
So the value of $ k $ is 1003 and hence the correct option is A. \[\]

Note:
 We can also alternatively solve by putting $ k=n+1 $ in the given condition. We know that we could divide all the terms by $ n $ and $ n-1 $ exists because $ n\ge 2 $ .We note that since $ n\in \mathsf{\mathbb{N}} $ the function $ nf\left( n \right) $ will also take inputs from the same domain as $ f\left( n \right) $ .We can also directly find the patterns of $ f\left( 2 \right),f\left( 3 \right),... $ and so on by directly evaluating from the given condition. We should always remember that the addition of a domain is a set function takes value and range is set to where the function returns value.