Summation of sequence terms

The question asks to compute or express a finite or infinite sum involving terms of a sequence defined by a recurrence relation.

jee-main 2025 Q84 View
Q84. Let $\lim _ { n \rightarrow \infty } \left( \frac { n } { \sqrt { n ^ { 4 } + 1 } } - \frac { 2 n } { \left( n ^ { 2 } + 1 \right) \sqrt { n ^ { 4 } + 1 } } + \frac { n } { \sqrt { n ^ { 4 } + 16 } } - \frac { 8 n } { \left( n ^ { 2 } + 4 \right) \sqrt { n ^ { 4 } + 16 } } + \ldots + \frac { n } { \sqrt { n ^ { 4 } + n ^ { 4 } } } - \frac { 2 n \cdot n ^ { 2 } } { \left( n ^ { 2 } + n ^ { 2 } \right) \sqrt { n ^ { 4 } + n ^ { 4 } } } \right)$ be $\frac { \pi } { k }$, using only the principal values of the inverse trigonometric functions. Then $\mathrm { k } ^ { 2 }$ is equal to $\_\_\_\_$
jee-main 2026 Q19 View
If $\mathbf{x}^{\mathbf{2}} + \mathbf{x} + \mathbf{1} = \mathbf{0}$,
then $\left(x + \frac{1}{x}\right)^{4} + \left(x^{2} + \frac{1}{x^{2}}\right)^{4} + \left(x^{3} + \frac{1}{x^{3}}\right)^{4} + \cdots + \left(x^{25} + \frac{1}{x^{25}}\right)^{4}$ is
jee-main 2026 Q21 View
If $\sum _ { \mathrm { k } = 1 } ^ { \mathrm { n } } \mathrm { a } _ { \mathrm { k } } = \alpha \mathrm { n } ^ { 2 } + \beta \mathrm { n }$ and $\mathrm { a } _ { 6 } = 7 \mathrm { a } _ { 1 } , \mathrm { a } _ { 10 } = 59$, then find the value of $\alpha + \beta$. (A) 6 (B) 5 (C) 10 (D) 8
jee-main 2026 Q34 View
Consider a sequence $729,81,9,1 , \ldots \ldots \quad 3 ^ { 6 } , 3 ^ { 4 } , 3 ^ { 2 } , 3 ^ { 0 } \ldots$
Let $\mathbf { P } _ { \mathbf { n } } =$ product of first $\mathbf { n }$ terms of the given sequence and $\sum _ { n = 1 } ^ { 40 } \left( P _ { n } \right) ^ { \frac { 1 } { n } } = \frac { 3 ^ { \alpha } - 1 } { 2 \times 3 ^ { \beta } }$. Then the value of $\alpha + \beta$ is
(A) 75
(B) 73
(C) 76
(D) 81
kyotsu-test 2013 QCourse2-IV-Q1 View
Let us define a sequence $\left\{ S _ { n } \right\}$ as
$$S _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { \sqrt { k } } \quad ( n = 1,2,3 , \cdots ) .$$
We are to find the following two limits:
$$\begin{aligned} & \lim _ { n \rightarrow \infty } S _ { n } , \\ & \lim _ { n \rightarrow \infty } \frac { S _ { 2 n } - S _ { n } } { \sqrt { n } } . \end{aligned}$$
(1) For each of A $\sim$ I in the following sentences, choose the appropriate answer from among (0) $\sim$ (9) at the bottom of this page.
Let us find $\lim _ { n \rightarrow \infty } S _ { n }$. Look at the function $y = \frac { 1 } { \sqrt { x } }$. We have
$$y ^ { \prime } = - \frac { \mathbf { A } } { 2 \sqrt { x ^ { \mathbf { B} } } } ,$$
and hence this function $y$ is $\square$ C . So, considering each interval $k \leqq x \leqq k + 1 ( k = 1,2 , \cdots , n )$, we obtain
$$\frac { 1 } { \sqrt { k } } \mathbf { D } \int _ { k } ^ { k + 1 } \frac { 1 } { \sqrt { x } } d x .$$
When we separately add the left-hand sides and the right-hand sides of this expression from $k = 1$ to $k = n$, we have
$$S _ { n } \mathbf { E } \int _ { \mathbf { F } } ^ { \mathbf { G } } \frac { 1 } { \sqrt { x } } d x = \mathbf { H } ( \sqrt { \square \mathbf { G } } - 1 )$$
and finally
$$\lim _ { n \rightarrow \infty } S _ { n } = \infty .$$
Choices: (0) $\infty$
(1) 1
(2) 2
(3) 3
(4) $n$
(5) $n + 1$ (6) $<$ (7) $>$ (8) monotonically increasing (9) monotonically decreasing
(2) For each of $\square$ J $\sim$ $\square$ P in the following, choose the appropriate answer from among (0) $\sim$ (9) below.
Let us find $\lim _ { n \rightarrow \infty } \frac { S _ { 2 n } - S _ { n } } { \sqrt { n } }$. Since
$$S _ { 2 n } - S _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { \sqrt { \mathbf { J } } } ,$$
we have from quadrature (mensuration) by parts that
$$\begin{aligned} \lim _ { n \rightarrow \infty } \frac { S _ { 2 n } - S _ { n } } { \sqrt { n } } & = \lim _ { n \rightarrow \infty } \frac { 1 } { \mathbf { K } } \sum _ { k = 1 } ^ { n } \frac { 1 } { \sqrt { \mathbf { L } + \frac { k } { n } } } \\ & = \int _ { \mathbf { M } } ^ { \mathbf { N } } \frac { 1 } { \sqrt { 1 + x } } d x \\ & = \mathbf { O } ( \sqrt { \mathbf { P } } - 1 ) . \end{aligned}$$
Choices: (0) 0
(1) 1
(2) 2
(3) $n - 1$
(4) $n$
(5) $n + 1$ (6) $n - k$ (7) $n + k$ (8) $n + k - 1$ (9) $n + k + 1$
mat 2016 Q5 View
5. For ALL APPLICANTS.
This question concerns the sum $s _ { n }$ defined by
$$s _ { n } = 2 + 8 + 24 + \cdots + n 2 ^ { n }$$
(i) Let $f ( n ) = ( A n + B ) 2 ^ { n } + C$ for constants $A , B$ and $C$ yet to be determined, and suppose $s _ { n } = f ( n )$ for all $n \geqslant 1$. By setting $n = 1,2,3$, find three equations that must be satisfied by $A , B$ and $C$.
(ii) Solve the equations from part (i) to obtain values for $A , B$ and $C$.
(iii) Using these values, show that if $s _ { k } = f ( k )$ for some $k \geqslant 1$ then $s _ { k + 1 } = f ( k + 1 )$.
You may now assume that $f ( n ) = s _ { n }$ for all $n \geqslant 1$.
(iv) Find simplified expressions for the following sums:
$$\begin{aligned} & t _ { n } = n + 2 ( n - 1 ) + 4 ( n - 2 ) + 8 ( n - 3 ) + \cdots + 2 ^ { n - 1 } 1 , \\ & u _ { n } = \frac { 1 } { 2 } + \frac { 2 } { 4 } + \frac { 3 } { 8 } + \cdots + \frac { n } { 2 ^ { n } } . \end{aligned}$$
(v) Find the sum
$$\sum _ { k = 1 } ^ { n } s _ { k }$$
If you require additional space please use the pages at the end of the booklet
mat 2017 Q2 View
2. For ALL APPLICANTS.
There is a unique real number $\alpha$ that satisfies the equation
$$\alpha ^ { 3 } + \alpha ^ { 2 } = 1$$
[You are not asked to prove this.]
(i) Show that $0 < \alpha < 1$.
(ii) Show that
$$\alpha ^ { 4 } = - 1 + \alpha + \alpha ^ { 2 }$$
(iii) Four functions of $\alpha$ are given in (a) to (d) below. In a similar manner to part (ii), each is equal to a quadratic expression
$$A + B \alpha + C \alpha ^ { 2 }$$
in $\alpha$, where $A , B , C$ are integers. (So in (ii) we found $A = - 1 , B = 1 , C = 1$.) You may assume in each case that the quadratic expression is unique.
In each case below find the quadratic expression in $\alpha$.
(a) $\alpha ^ { - 1 }$.
(b) The infinite sum
$$1 - \alpha + \alpha ^ { 2 } - \alpha ^ { 3 } + \alpha ^ { 4 } - \alpha ^ { 5 } + \cdots$$
(c) $( 1 - \alpha ) ^ { - 1 }$.
(d) The infinite product
$$( 1 + \alpha ) \left( 1 + \alpha ^ { 2 } \right) \left( 1 + \alpha ^ { 4 } \right) \left( 1 + \alpha ^ { 8 } \right) \left( 1 + \alpha ^ { 16 } \right) \cdots$$
taiwan-gsat 2023 Q9 5 marks View
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
tmua None Q16 View
16. The sequence $a _ { n }$ is given by the rule:
$$\begin{aligned} a _ { 1 } & = 2 \\ a _ { n + 1 } & = a _ { n } + ( - 1 ) ^ { n } \text { for } n \geq 1 \end{aligned}$$
What is
$$\sum _ { n = 1 } ^ { 100 } a _ { n }$$
A 150
B 250
C - 4750
D 5150
E $\quad 4 \left( 1 - \left( \frac { 1 } { 2 } \right) ^ { 100 } \right)$ F $\quad 4 \left( \left( \frac { 3 } { 2 } \right) ^ { 100 } - 1 \right)$
turkey-yks 2019 Q16 View
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