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

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

Explanation

Solution

To solve this question, we will first use the formula of f which is given by f(1)+2f(2)+....+nf(n)=n(n+1)f(n).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(999)=1k.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(1)+2f(2)+3f(3)....+nf(n)=n(n+1)f(n).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).
⇒f(1)+2f(2)+3f(3)....+nf(n)=n(n+1)f(n).....(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(1)+2f(2)+3f(3)....+nf(n)+(n+1)f(n+1)=(n+1)(n+2)f(n+1).....(ii)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,
⇒(n+1)f(n+1)=(n+1)(n+2)f(n+1)−n(n+1)f(n)\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)
⇒(n+1)(n+2)f(n+1)−(n+1)f(n+1)=n(n+1)f(n)\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)
⇒(n+1)2f(n+1)=n(n+1)f(n)\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,
⇒(n+1)f(n+1)=nf(n)\Rightarrow \left( n+1 \right)f\left( n+1 \right)=nf\left( n \right)
Now for any n, we have, (n+1)f(n+1)=nf(n).\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.
⇒kf(k)=C\Rightarrow kf\left( k \right)=C where C is a constant.
When k = 1,
⇒1f(1)=C\Rightarrow 1f\left( 1 \right)=C
Now, f(999)=1k,f\left( 999 \right)=\dfrac{1}{k}, and we have,
kf(k)=Ckf\left( k \right)=C
⇒f(k)=Ck\Rightarrow f\left( k \right)=\dfrac{C}{k}
⇒f(k)=1k\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(999)=1kf\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.