Question

The value of f(999) is \[\dfrac{1}{k},\] where k equals
(a) 999
(b) 1000
(c) 1998
(d) 2000

Answer 1

Hint: To solve this question, we will first use the formula of f which is given by \[f\left( 1 \right)+2f\left( 2 \right)+....+nf\left( n \right)=n\left( n+1 \right)f\left( n \right).\] Use n = n +1 to get the next equation and subtract them to get the value of k f(k) and finally get k.

Complete step by step answer:
We are given that, \[f\left( 999 \right)=\dfrac{1}{k}.\]
We will be making use of the summation of series. We know that we have a formula of function f given as \[f\left( 1 \right)+2f\left( 2 \right)+3f\left( 3 \right)....+nf\left( n \right)=n\left( n+1 \right)f\left( n \right).\] Let us consider this as equation (i).
\[\Rightarrow f\left( 1 \right)+2f\left( 2 \right)+3f\left( 3 \right)....+nf\left( n \right)=n\left( n+1 \right)f\left( n \right).....\left( i \right)\]
Let us use n = n +1 in the above equation, then we have,
\[f\left( 1 \right)+2f\left( 2 \right)+3f\left( 3 \right)....+nf\left( n \right)+\left( n+1 \right)f\left( n+1 \right)=\left( n+1 \right)\left( n+2 \right)f\left( n+1 \right).....\left( ii \right)\]
Now let us subtract equation (i) and (ii). Do that by subtracting the left-hand side of both equations by the left-hand side and right-hand side. Doing so, we will get,
\[\Rightarrow \left( n+1 \right)f\left( n+1 \right)=\left( n+1 \right)\left( n+2 \right)f\left( n+1 \right)-n\left( n+1 \right)f\left( n \right)\]
\[\Rightarrow \left( n+1 \right)\left( n+2 \right)f\left( n+1 \right)-\left( n+1 \right)f\left( n+1 \right)=n\left( n+1 \right)f\left( n \right)\]
\[\Rightarrow f\left( n+1 \right)\left\{ \left( n+1 \right)\left( n+2 \right)-\left( n+1 \right) \right\}=n\left( n+1 \right)f\left( n \right)\]
\[\Rightarrow \left( n+1 \right)f\left( n+1 \right)\left\{ n+1 \right\}=n\left( n+1 \right)f\left( n \right)\]
\[\Rightarrow {{\left( n+1 \right)}^{2}}f\left( n+1 \right)=n\left( n+1 \right)f\left( n \right)\]
Cancelling (n + 1) on both the sides of the equation, we have,
\[\Rightarrow \left( n+1 \right)f\left( n+1 \right)=nf\left( n \right)\]
Now for any n, we have, \[\left( n+1 \right)f\left( n+1 \right)=nf\left( n \right).\] This type of situation is only possible if the value of n f(n) is a constant. Let the constant be C, then n f(n) = C.
\[\Rightarrow kf\left( k \right)=C\] where C is a constant.
When k = 1,
\[\Rightarrow 1f\left( 1 \right)=C\]
Now, \[f\left( 999 \right)=\dfrac{1}{k},\] and we have,
\[kf\left( k \right)=C\]
\[\Rightarrow f\left( k \right)=\dfrac{C}{k}\]
\[\Rightarrow f\left( k \right)=\dfrac{1}{k}\] when C = 1.
And, k = 999 therefore, the value of k = 999.

So, the correct answer is “Option a”.

Note: One key point to note here in this question is that we have obtained the value of C using the given \[f\left( 999 \right)=\dfrac{1}{k}\] value. Always in such types of questions, try to obtain the value of the given term so that the value of the new value can be easily determined.