(A) Given the recurrence relation $f(x)=f(x-2)+f(x-3)$ with initial values $f(0)=0, f(1)=1, f(2)=2$.We calculate the subsequent values step by step:$f

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) Given the recurrence relation $f(x)=f(x-2)+f(x-3)$ with initial values $f(0)=0, f(1)=1, f(2)=2$.We calculate the subsequent values step by step:$f(3)=f(1)+f(0)=1+0=1$$f(4)=f(2)+f(1)=2+1=3$$f(5)=f(3)+f(2)=1+2=3$$f(6)=f(4)+f(3)=3+1=4$$f(7)=f(5)+f(4)=3+3=6$$f(8)=f(6)+f(5)=4+3=7$$f(9)=f(7)+f(6)=6+4=10$$f(10)=f(8)+f(7)=7+6=13$

Class 11 Mathematics Sequences and Series nth term of special series, Sum to n terms and Infinite number of terms

Easy

If $f(0)=0, f(1)=1, f(2)=2$ and $f(x)=f(x-2)+f(x-3)$ for $x=3, 4, 5, \ldots$,then $f(10)=$

AP EAMCET 2023 All AP EAMCET

A
$13$
B
$9$
C
$11$
D
$10$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Sequences and Series

Subtopic

nth term of special series, Sum to n terms and Infinite number of terms

Previous Year

AP EAMCET 2023 Paper All AP EAMCET years →

The sum of the series $1 + 3x + 6x^2 + 10x^3 + \dots \infty$ is

Medium

View Solution

The sum to $n$ terms of the series $1^2 + (1^2 + 3^2) + (1^2 + 3^2 + 5^2) + \dots$ is

Difficult

View Solution

$\sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots+(2 k-1)}$ is equal to (in $.5$)

DifficultAP EAMCET 2004

View Solution

For $\frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$ to exceed $1.01$,the maximum value of $n$ is

KVPY 2012

View Solution

What is the sum of the series $1^3 - 2^3 + 3^3 - 4^3 + 5^3 - 6^3 + 7^3 - 8^3 + 9^3$?

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If $f(0)=0, f(1)=1, f(2)=2$ and $f(x)=f(x-2)+f(x-3)$ for $x=3, 4, 5, \ldots$,then $f(10)=$

Similar Questions

The sum of the series $1 + 3x + 6x^2 + 10x^3 + \dots \infty$ is

The sum to $n$ terms of the series $1^2 + (1^2 + 3^2) + (1^2 + 3^2 + 5^2) + \dots$ is

$\sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots+(2 k-1)}$ is equal to (in $.5$)

For $\frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$ to exceed $1.01$,the maximum value of $n$ is

What is the sum of the series $1^3 - 2^3 + 3^3 - 4^3 + 5^3 - 6^3 + 7^3 - 8^3 + 9^3$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module