(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

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(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

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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
$10$

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

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Find the sum of the following series up to $n$ terms:
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DifficultJEE MAIN 2022

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If $S_n$ is the sum of the first $n$ terms of the series $1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots$,then for even $n$,$S_n$ is equal to:

EasyTS EAMCET 2020

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The sum of the series $\frac{1}{1-3 \cdot 1^2+1^4} + \frac{2}{1-3 \cdot 2^2+2^4} + \frac{3}{1-3 \cdot 3^2+3^4} + \ldots$ up to $10$ terms is

DifficultJEE MAIN 2024

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The sum of $n$ terms of the following series $1 + (1 + x) + (1 + x + x^2) + \dots$ will be

MediumIIT 1962

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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 $f(10)=$

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Find the sum of the following series up to $n$ terms:$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$

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Find the sum of the following series up to $n$ terms:
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