FunctionsmediumFree

Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Let ff be defined on the natural numbers as follows: f(1)=1f(1) = 1 and for n>1n > 1, f(n)=f(f(n−1))+f(n−f(n−1))f(n) = f(f(n-1)) + f(n - f(n-1)). then the value of 130∑r=120f(r)\dfrac{1}{30}\displaystyle\sum_{r=1}^{20} f(r).
Solution
Answer: 7
Step 1: Show f(n)=nf(n) = n for all nn Using the recurrence with f(n−1)=n−1f(n-1) = n-1:
f(n)=f(n−1)+f(n−(n−1))=f(n−1)+f(1)=(n−1)+1=nf(n) = f(n-1) + f(n-(n-1)) = f(n-1) + f(1) = (n-1)+1 = n
 Verified: f(2)=f(f(1))+f(2−f(1))=f(1)+f(1)=2f(2) = f(f(1))+f(2-f(1)) = f(1)+f(1) = 2, f(3)=f(2)+f(1)=3f(3) = f(2)+f(1) = 3, and so on. Step 2: Evaluate the sum
130∑r=120f(r)=130∑r=120r=130⋅20×212=21030=7\frac{1}{30}\sum_{r=1}^{20} f(r) = \frac{1}{30}\sum_{r=1}^{20} r = \frac{1}{30}\cdot\frac{20 \times 21}{2} = \frac{210}{30} = 7
 Answer: 7
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Functions · easy
If [x] denotes the integral part of x , then domain of the function f(x)= 3-x (x-1)(x-2)(…
Functions · easy
The range of the function y=[x 2 ]-[x] 2 , x [0,2] (where [ ] denotes the integral part)…
Functions · easy
The number of integers in the domain of f(x) = 1 -1 x .
Functions · hard
If f:(0, /n) R defined by f(x)= k=1 n [1+ kx] , where [x] denotes the integral part of x…

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath