Sequences and series, recurrence and convergence

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jee-main 2024 Q61 Multiple-choice on sequence properties
Let $\alpha , \beta$ be the distinct roots of the equation $x ^ { 2 } - \left( t ^ { 2 } - 5 t + 6 \right) x + 1 = 0 , t \in \mathbb { R }$ and $a _ { n } = \alpha ^ { n } + \beta ^ { n }$. Then the minimum value of $\frac { a _ { 2023 } + a _ { 2025 } } { a _ { 2024 } }$ is
(1) $- 1 / 4$
(2) $- 1 / 4$
(3) $- 1 / 2$
(4) $1 / 4$
jee-main 2024 Q61 Multiple-choice on sequence properties
If $\alpha , \beta$ are the roots of the equation, $\mathrm { x } ^ { 2 } - \mathrm { x } - 1 = 0$ and $\mathrm { S } _ { \mathrm { n } } = 2023 \alpha ^ { \mathrm { n } } + 2024 \beta ^ { \mathrm { n } }$, then
(1) $2 \quad \mathrm {~S} _ { 12 } = \mathrm { S } _ { 11 } + \mathrm { S } _ { 10 }$
(2) $\mathrm { S } _ { 12 } = \mathrm { S } _ { 11 } + \mathrm { S } _ { 10 }$
(3) $2 \mathrm {~S} _ { 11 } = \mathrm { S } _ { 12 } + \mathrm { S } _ { 10 }$
(4) $\mathrm { S } _ { 11 } = \mathrm { S } _ { 10 } + \mathrm { S } _ { 12 }$
jee-main 2024 Q61 Multiple-choice on sequence properties
Let $\alpha , \beta ; \alpha > \beta$, be the roots of the equation $x ^ { 2 } - \sqrt { 2 } x - \sqrt { 3 } = 0$. Let $\mathrm { P } _ { n } = \alpha ^ { n } - \beta ^ { n } , n \in \mathrm {~N}$. Then $( 11 \sqrt { 3 } - 10 \sqrt { 2 } ) \mathrm { P } _ { 10 } + ( 11 \sqrt { 2 } + 10 ) \mathrm { P } _ { 11 } - 11 \mathrm { P } _ { 12 }$ is equal to
(1) $10 \sqrt { 3 } \mathrm { P } _ { 9 }$
(2) $11 \sqrt { 3 } P _ { 9 }$
(3) $10 \sqrt { 2 } \mathrm { P } _ { 9 }$
(4) $11 \sqrt { 2 } \mathrm { P } _ { 9 }$
jee-main 2024 Q68 Convergence proof and limit determination
$\lim _ { n \rightarrow \infty } \frac { \left( 1 ^ { 2 } - 1 \right) ( n - 1 ) + \left( 2 ^ { 2 } - 2 \right) ( n - 2 ) + \cdots + \left( ( n - 1 ) ^ { 2 } - ( n - 1 ) \right) \cdot 1 } { \left( 1 ^ { 3 } + 2 ^ { 3 } + \cdots \cdots + n ^ { 3 } \right) - \left( 1 ^ { 2 } + 2 ^ { 2 } + \cdots \cdots + n ^ { 2 } \right) }$ is equal to:
(1) $\frac { 2 } { 3 }$
(2) $\frac { 1 } { 3 }$
(3) $\frac { 3 } { 4 }$
(4) $\frac { 1 } { 2 }$
jee-main 2024 Q81 Direct term computation from recurrence
Let $\alpha , \beta$ be roots of $x ^ { 2 } + \sqrt { 2 } x - 8 = 0$. If $\mathrm { U } _ { \mathrm { n } } = \alpha ^ { \mathrm { n } } + \beta ^ { n }$, then $\frac { \mathrm { U } _ { 10 } + \sqrt { 2 } \mathrm { U } _ { 9 } } { 2 \mathrm { U } _ { 8 } }$ is equal to $\_\_\_\_$
jee-main 2024 Q82 Series convergence and power series analysis
If $1 + \frac { \sqrt { 3 } - \sqrt { 2 } } { 2 \sqrt { 3 } } + \frac { 5 - 2 \sqrt { 6 } } { 18 } + \frac { 9 \sqrt { 3 } - 11 \sqrt { 2 } } { 36 \sqrt { 3 } } + \frac { 49 - 20 \sqrt { 6 } } { 180 } + \ldots$ upto $\infty = 2 + \left( \sqrt { \frac { b } { a } } + 1 \right) \log _ { e } \left( \frac { a } { b } \right)$, where a and b are integers with $\operatorname { gcd } ( \mathrm { a } , \mathrm { b } ) = 1$, then $11 \mathrm { a } + 18 \mathrm {~b}$ is equal to $\_\_\_\_$
jee-main 2025 Q16 Series convergence and power series analysis
The value of $\lim_{n \rightarrow \infty}\left(\sum_{k=1}^{n} \frac{k^3 + 6k^2 + 11k + 5}{(k+3)!}\right)$ is:
(1) $4/3$
(2) $2$
(3) $7/3$
(4) $5/3$
jee-main 2025 Q18 Summation of sequence terms
Let $\left\langle a _ { \mathrm { n } } \right\rangle$ be a sequence such that $a _ { 0 } = 0 , a _ { 1 } = \frac { 1 } { 2 }$ and $2 a _ { \mathrm { n } + 2 } = 5 a _ { \mathrm { n } + 1 } - 3 a _ { \mathrm { n } } , \mathrm { n } = 0,1,2,3 , \ldots$ Then $\sum _ { \mathrm { k } = 1 } ^ { 100 } a _ { k }$ is equal to
(1) $3 a _ { 99 } - 100$
(2) $3 \mathrm { a } _ { 100 } - 100$
(3) $3 a _ { 99 } + 100$
(4) $3 \mathrm { a } _ { 100 } + 100$
kyotsu-test 2016 QCourse2-II Auxiliary sequence transformation
Consider a sequence of positive numbers $a _ { 1 } , a _ { 2 } , a _ { 3 } , \cdots$ which satisfies
$$\begin{aligned} a _ { 1 } & = 1 , \quad a _ { 2 } = 10 , \\ \left( a _ { n } \right) ^ { 2 } a _ { n - 2 } & = \left( a _ { n - 1 } \right) ^ { 3 } \quad ( n = 3,4 , \cdots ) . \end{aligned}$$
We are to find $\lim _ { n \rightarrow \infty } a _ { n }$.
By finding the common logarithm of both sides of (1), we obtain
$$\mathbf { A } \log _ { 10 } a _ { n } + \log _ { 10 } a _ { n - 2 } = \mathbf { B } \log _ { 10 } a _ { n - 1 } .$$
When we set $b _ { n } = \log _ { 10 } a _ { n }$ $(n = 1,2 , \cdots)$, this equality is expressed as
$$\mathbf { A } b _ { n } + b _ { n - 2 } = \mathbf { B } b _ { n - 1 } .$$
By transforming (2), we have
$$b _ { n } - b _ { n - 1 } = \frac { 1 } { \mathbf { C } } \left( b _ { n - 1 } - b _ { n - 2 } \right) \quad ( n = 3,4 , \cdots ) ,$$
which gives
$$b _ { n } - b _ { n - 1 } = \left( \frac { 1 } { \mathbf { C } } \right) ^ { n - \mathbf { D } } \left( b _ { 2 } - b _ { 1 } \right) \quad ( n = 2,3 , \cdots ) .$$
Since $b _ { 1 } = \mathbf { E }$ and $b _ { 2 } = \mathbf { F }$, from (3) we get
$$b _ { n } = \sum _ { k = 2 } ^ { n } \left( \frac { 1 } { \mathbf { C } } \right) ^ { k - \mathbf { G } }$$
and hence
$$b _ { n } = \mathbf { H } - \left( \frac { 1 } { \mathbf { C } } \right) ^ { n - \mathbf{I} }$$
Finally, we obtain
$$\lim _ { n \rightarrow \infty } a _ { n } = \mathbf { J K L } .$$
kyotsu-test 2017 QCourse2-II-Q1 Auxiliary sequence transformation
Q1 We are to find the general term $a _ { n }$ of the sequence $\left\{ a _ { n } \right\}$ which is determined by the recurrence formula
$$a _ { 1 } = 18 , \quad a _ { n + 1 } - 12 a _ { n } + 3 ^ { n + 2 } = 0 \quad ( n = 1,2,3 , \cdots ) .$$
When we define a sequence $\left\{ b _ { n } \right\}$ by
$$b _ { n } = \frac { a _ { n } } { \mathbf{A}^n } \quad ( n = 1,2,3 , \cdots ) ,$$
$\left\{ b _ { n } \right\}$ satisfies
$$b _ { 1 } = \mathbf { B } , \quad b _ { n + 1 } - \mathbf { C } \; b _ { n } + \mathbf { D } = 0 \quad ( n = 1,2,3 , \cdots ) .$$
This recurrence formula can be transformed into
$$b _ { n + 1 } - \mathbf { E } = \mathbf{F} ( b_n - \mathbf{E} )$$
Next, when we define a sequence $\left\{ c _ { n } \right\}$ by
$$c _ { n } = b _ { n } - \mathbf { E } \quad ( n = 1,2,3 , \cdots ) ,$$
$\left\{ c _ { n } \right\}$ is a geometric progression such that the first term is $\mathbf{G}$ and the common ratio is $\mathbf{H}$.
Hence we have
$$a _ { n } = \mathbf { I } ^ { n } \left( \mathbf { J } \cdot \mathbf { K } ^ { n - 1 } + \mathbf { L } \right) \quad ( n = 1,2,3 , \cdots ) .$$
taiwan-gsat 2020 Q5 8 marks True/false or conceptual reasoning about sequences
For a real number $a$, let $[a]$ denote the greatest integer not exceeding $a$. For example: $[1.2] = [\sqrt{2}] = 1$, $[-1.2] = -2$. Consider the irrational number $\theta = \sqrt{10001}$. Select the correct options.
(1) $a - 1 < [a] \leq a$ holds for all real numbers $a$
(2) The sequence $b_{n} = \frac{[n\theta]}{n}$ diverges, where $n$ is a positive integer
(3) The sequence $c_{n} = \frac{[-n\theta]}{n}$ diverges, where $n$ is a positive integer
(4) The sequence $d_{n} = n\left[\frac{\theta}{n}\right]$ diverges, where $n$ is a positive integer
(5) The sequence $e_{n} = n\left[\frac{-\theta}{n}\right]$ diverges, where $n$ is a positive integer
taiwan-gsat 2021 Q6 8 marks Multiple-choice on sequence properties
Given a real number sequence $\left\langle a _ { n } \right\rangle$ satisfying $a _ { 1 } = 1 , a _ { n + 1 } = \frac { 2 n + 1 } { 2 n - 1 } a _ { n } , n$ is a positive integer. Select the correct options.
(1) $a _ { 2 } = 3$
(2) $a _ { 4 } = 9$
(3) $\left\langle a _ { n } \right\rangle$ is a geometric sequence
(4) $\sum _ { n = 1 } ^ { 20 } a _ { n } = 400$
(5) $\lim _ { n \rightarrow \infty } \frac { a _ { n } } { n } = 2$
taiwan-gsat 2022 Q8 8 marks Convergence proof and limit determination
Suppose two sequences $\langle a_n \rangle$ and $\langle b_n \rangle$ satisfy $b_n + \frac{4n-1}{n} < a_n < 3b_n$ for all positive integers $n$. Given that $\lim_{n \to \infty} a_n = 6$, select the correct options.
(1) $b_n < 6 - \frac{4n-1}{n}$
(2) $b_n > \frac{4n-1}{2n}$
(3) The sequence $\langle b_n \rangle$ may diverge
(4) $a_{10000} < 6.1$
(5) $a_{10000} > 5.9$
taiwan-gsat 2023 Q9 5 marks Summation of sequence terms
Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be a geometric sequence with first term 3 and common ratio $3\sqrt{3}$. Select the number of terms $n$ that satisfy the inequality
$$\log_{3} a_{1} - \log_{3} a_{2} + \log_{3} a_{3} - \log_{3} a_{4} + \ldots + (-1)^{n+1} \log_{3} a_{n} > 18$$
among the possible options.
(1) 23
(2) 24
(3) 25
(4) 26
(5) 27
taiwan-gsat 2024 Q11 6 marks Closed-form expression derivation
Let the real numbers $a_{1}, a_{2}, \ldots, a_{9}$ form an arithmetic sequence with common difference 2, where $a_{1} \neq 0$ and $a_{3} > 0$. If $\log_{2} a_{3}$, $\log_{2} b$, $\log_{2} a_{9}$ form an arithmetic sequence in order, where $b$ is one of $a_{4}, a_{5}, a_{6}, a_{7}, a_{8}$, then $a_{9} = $ . (Express as a simplified fraction)
taiwan-gsat 2025 Q5 8 marks True/false or conceptual reasoning about sequences
There is a real number sequence $\left\langle a_{n} \right\rangle$, where $a_{n} = \cos\left(n\pi - \frac{\pi}{6}\right)$, and $n$ is a positive integer. Select the correct options.
(1) $a_{1} = -\frac{1}{2}$
(2) $a_{2} = a_{3}$
(3) $a_{4} = a_{24}$
(4) $\left\langle a_{n} \right\rangle$ is a convergent sequence, and $\lim_{n \rightarrow \infty} a_{n} < 1$
(5) $\sum_{n=1}^{\infty} \left(a_{n}\right)^{n} = 3 - 2\sqrt{3}$
taiwan-gsat 2025 Q7 5 marks Auxiliary sequence transformation
A sequence $< a _ { n } >$ satisfies $3 a _ { n + 1 } = a _ { n } + n$ (for all positive integers $n$) and $a _ { 1 } = 2$. Let the sequence $< b _ { n } >$ satisfy $b _ { n } = a _ { n } - \frac { n } { 2 } + \frac { 3 } { 4 }$. Select the correct options.
(1) $a _ { 2 } = 2$
(2) $b _ { 2 } = \frac { 3 } { 4 }$
(3) The sequence $< b _ { n } >$ is a geometric sequence with common ratio $\frac { 2 } { 3 }$
(4) For any positive integer $n$, $3 ^ { n } a _ { n }$ is always a positive integer
(5) $b _ { 10 } < 10 ^ { - 4 }$
todai-math 2025 Q3 Closed-form expression derivation
A coin with the probability of coming up heads equal to $p$ is tossed $n$ times. Let $a _ { n }$ be the probability that the coin never comes up heads twice in a row, and let $b _ { n }$ be the probability that the coin never comes up heads three times in a row. Here, let $a _ { 1 } = 1$ and $b _ { 1 } = b _ { 2 } = 1$. Answer the following questions.
(1) Obtain $a _ { 2 }$ as a function of $p$.
(2) When $n \geq 3$, describe $a _ { n }$ by using $p , a _ { n - 1 }$ and $a _ { n - 2 }$.
(3) When $p = \frac { 2 } { 3 }$, obtain all the pairs $( \alpha , \beta )$ of real numbers that satisfy the following recursion:
$$a _ { n } + \alpha a _ { n - 1 } = \beta \left( a _ { n - 1 } + \alpha a _ { n - 2 } \right)$$
(4) When $p = \frac { 2 } { 3 }$, obtain $a _ { n }$ as a function of $n$.
(5) Obtain $b _ { 3 }$ as a function of $p$.
(6) When $n \geq 4$, describe $b _ { n }$ by using $p , b _ { n - 1 } , b _ { n - 2 }$ and $b _ { n - 3 }$.
(7) When $p = \frac { 3 } { 4 }$, show that the following equation holds for any positive integer $n$:
$$\begin{aligned} b _ { n } = & \frac { 9 } { 8 } \left( \frac { 3 } { 4 } \right) ^ { n - 1 } - \frac { ( - 1 ) ^ { n - 1 } } { 8 } \left( \frac { \sqrt { 3 } } { 4 } \right) ^ { n - 1 } \cos ( ( n - 1 ) \theta ) \\ & - \frac { \sqrt { 2 } \mathrm { i } } { 8 } \left\{ \left( \frac { - 1 + \sqrt { 2 } \mathrm { i } } { 4 } \right) ^ { n - 1 } - \left( \frac { - 1 - \sqrt { 2 } \mathrm { i } } { 4 } \right) ^ { n - 1 } \right\} \end{aligned}$$
where i is the imaginary unit and $\theta$ is the angle that satisfies $\cos \theta = \frac { 1 } { \sqrt { 3 } }$ and $\sin \theta = \frac { \sqrt { 2 } } { \sqrt { 3 } }$.
turkey-yks 2011 Q33 Direct term computation from recurrence
The function f satisfies the equation
$$f ( n ) = 2 \cdot f ( n - 1 ) + 1$$
for integers $n \geq 1$. Given that $f ( 0 ) = 1$, what is $f ( 2 )$?
A) 8
B) 7
C) 6
D) 5
E) 4
turkey-yks 2011 Q34 Direct term computation from recurrence
The sequence $(a _ { k })$ is defined as
$$\begin{aligned} & a _ { 1 } = 40 \\ & a _ { k + 1 } = a _ { k } - k \quad ( k = 1,2,3 , \ldots ) \end{aligned}$$
Accordingly, what is the term $\mathrm { a } _ { 8 }$?
A) 4
B) 7
C) 12
D) 15
E) 19
turkey-yks 2013 Q32 Direct term computation from recurrence
Let $a _ { 1 } , a _ { 2 }$ be real numbers. The sequence $\left( a _ { n } \right)$ satisfies the relation
$$a _ { n + 2 } = a _ { n + 1 } + a _ { n } \quad ( n = 1,2 , \cdots )$$
Given that $a _ { 8 } = 6$, what is the sum $a _ { 6 } + a _ { 9 }$?
A) 9
B) 10
C) 12
D) 15
E) 16
turkey-yks 2019 Q16 Summation of sequence terms
For a sequence $a_n$ where the sum of any three consecutive terms is equal to each other,
$$a _ { 2 } + a _ { 3 } = a _ { 4 } = 2$$
equality is satisfied.
Accordingly, $$a _ { 1 } + a _ { 2 } + \ldots + a _ { 25 }$$
what is the result of the sum?
A) 34
B) 35
C) 36
D) 37
E) 38
turkey-yks 2020 Q23 Direct term computation from recurrence
The sequence $(a_n)$ of real numbers satisfies for every positive integer $n$
$$a_{n+1} = a_n + \frac{(-1)^n \cdot a_n}{2}$$
the equality. If $a_5 = 18$, what is $a_1$?
A) 4
B) 8
C) 16
D) 32
E) 64
turkey-yks 2023 Q25 Direct term computation from recurrence
The sequence $(a_n)$ of real numbers satisfies for every positive integer $n$
$$a_{n} + (-1)^{n} \cdot a_{n+1} = 2^{n}$$
If $a_{1} = 0$, what is the sum $a_{3} + a_{4} + a_{5} + a_{6}$?
A) 6 B) 8 C) 12 D) 16 E) 20
turkey-yks 2024 Q3 Applied/contextual sequence problem
Selma designs a toy with identical clips in yellow and blue colors. In step 1, she places one clip on the ground. In each subsequent step, she attaches one clip to each blue part of all the clips she placed in the previous step as shown in the figure, and moves to the next step. Selma completes the first 3 steps of this toy using 7 clips.
Accordingly, after Selma completes step 12, how many more clips has she used in total compared to after completing step 10?
A) $3 \cdot 2^{10}$
B) $3 \cdot 2^{11}$
C) $7 \cdot 2^{9}$
D) $7 \cdot 2^{10}$
E) $2^{11}$