Sequence and Integers | AIME I, 2007 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.

Sequence and Integers - AIME I, 2007

A sequence is defined over non negetive integral indexes in the following way $a_0=a_1=3$, $ a_{n+1}a_{n-1}=a_n^{2}+2007$, find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$

is 107
is 224
is 840
cannot be determined from the given information

Key Concepts

Sequence

Inequalities

Integers

Check the Answer

Answer: is 224.

AIME I, 2007, Question 14

Elementary Number Theory by David Burton

Try with Hints

$a_{n+1}a_{n-1}$=$a_{n}^{2}+2007$ then $a_{n-1}^{2} +2007 =a_{n}a_{n-2}$ adding these $\frac{a_{n-1}+a_{n+1}}{a_{n}}$=$\frac{a_{n}+a_{n-2}}{a_{n-1}}$, let $b_{j}$=$\frac{a_{j}}{a_{j-1}}$ then $b_{n+1} + \frac{1}{b_{n}}$=$b_{n}+\frac{1}{b_{n-1}}$ then $b_{2007} + \frac{1}{b_{2006}}$=$b_{3}+\frac{1}{b_{2}}$=225

here $\frac{a_{2007}a_{2005}}{a_{2006}a_{2005}}$=$\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$ then $b_{2007}$=$\frac{a_{2007}}{a_{2006}}$=$\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$$ \gt $$\frac{a_{2006}}{a_{2005}}$=$b_{2006}$

then $b_{2007}+\frac{1}{b_{2007}} \lt b_{2007}+\frac{1}{b_{2006}}$=225 which is small less such that all $b_{j}$ s are greater than 1 then $\frac{a_{2006}^{2}+ a_{2007}^{2}}{a_{2006}a_{2007}}$=$b_{2007}+\frac{1}{b_{2007}}$=224.

GCD and Sequence | AIME I, 1985 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on GCD and Sequence.

GCD and Sequence - AIME I, 1985

The numbers in the sequence 101, 104,109,116,.....are of the form $a_n=100+n^{2}$ where n=1,2,3,-------, for each n, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$, find the maximum value of $d_n$ as n ranges through the positive integers.

is 107
is 401
is 840
cannot be determined from the given information

Key Concepts

GCD

Sequence

Integers

Check the Answer

Answer: is 401.

AIME I, 1985, Question 13

Elementary Number Theory by David Burton

Try with Hints

$a_n=100+n^{2}$ $a_{n+1}=100+(n+1)^{2}=100 + n^{2} +2n +1$ and $a_{n+1}-a_{n}=2n +1$

$d_{n}|(2n+1)$ and $d_{n}|(100 +n^{2})$ then $d_{n}|[(100+n^{2})-100(2n+1)]$ then $d_{n}|(n^{2}-200n)$

here $n^{2} -200n=0$ then n=200 then $d_{n}$=2n+1=2(200)+1=401.

Problem on Series | SMO, 2009 | Problem No. 25

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Problem on Series.

Problem on Series | SMO Test

Given that $x+(1+x)^2+(1+x)^3+.........(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n$ where each $ a_r$ is an integer , $r = 0,1,2,...,n$

Find the value of n such that $a_0 +a_1 +a_2+a_3 +...........+a_{n-2}+a_{n-1} = 60 -\frac{n(n-1)}{2}$?

Key Concepts

Series Problem

Algebra

Check the Answer

Answer : 5

Singapore Mathematics Olympiad, 2009

Challenges and Thrills - Pre - college Mathematics

Try with Hints

If you got stuck in this sum then we can try by understanding the pattern we are using here.If we assume x = 1 thenn the expression will be

$x+(1+x)^2+(1+x)^3+.........(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n$

The right hand side of this equation will be

$a_0 +a_1 + a_2 +a_3+..................+ a_n$

and the left hand side of the given equation be like

$ 1 + 2^2 + 2^ 3 + ......+2^n$

$ 1 + 2^2 + 2^ 3 + ......+2^n$ = $a_0 +a_1 + a_2 +a_3+..................+ a_n$

Try the rest of the sum..............

In the next hint we continue from the previous hint:

so the expression , $ 1 + 2^2 + 2^ 3 + ......+2^n$ = $2^{n+1} - 3$

Again , $a_1 = 1+2+3+..........+n = \frac {n(n+1)}{2}$

so $a_n = 1$

Now we have almost reach the final step.I am not showing it now.Try first.........

Now coming back to the last step :

$ 60 - \frac {n(n+1)}{2} + \frac {n(n+1}{2} + 1 = (2^{n+1} - 3$

n = 5 (Answer)

Problem on Series and Sequences | SMO, 2012 | Problem 23

Try this beautiful problem from Singapore Mathematics Olympiad, 2012 based on Series and Sequences.

Problem on Series and Sequences (SMO Test)

For each positive integer $n \geq 1$ , we define the recursive relation given by $a_{n+1} = \frac {1}{1+a_{n}} $.

Suppose that $a_{1} = a_{2012}$.Find the sum of the squares of all

possible values of $a_{1}$.

Key Concepts

Series and Sequence

Functional Equation

Recursive Relation

Check the Answer

Answer: 3

Singapore Mathematics Olympiad

Challenges and Thrills - Pre - College Mathematics

Try with Hints

If you got stuck start from here :

At first we have to understand the sequence it is following

Given that : $a_{n+1} = \frac {1}{1+a_{n}} $

Let $a_{1} = a$

so , $a_{2} = \frac {1}{1 + a_{1}}$ = $\frac {1}{1 + a}$

Again, $a_{3} = \frac {1}{1+a_{2}} = \frac {1+a}{2+a} $

For , $a_{4} = \frac {1}{1+a_{3}} = \frac {2+a}{3+2a} $

And , $a_{5} = \frac {1}{1+a_{4}} = \frac {3+2a}{5+3a} $

and so on........

Try to do the rest ..................................

Looking at the previous hint ..................

In general we can say .................

$a_{n} = \frac {F_{n} + F_{n-1}a}{F_{n+1} +F_{n}a}$

Where $F_{1} = 0 , F_{ 2} = 1$ and $F_{n+1} = F_{n} $ for all value of $n\geq 1$

Try to do the rest .......

Here is the rest of the solution,

If $a_{2012} = \frac {F_{2012}+F_{2011}a}{F_{2013} + F{2012}a} = a $

Then $(a^2+a-1 )F_{2012} = 0$

Since $F_{2012}>0$ we have $a^2 +a -1 = 0$ ............................(1)

Assume x and y are the two roots of the $eq^n (1)$, then

$x^2 + y^2 = (x+y)^2 -2xy = (-1)^2 - 2(-1) = 3$ (Answer)

Trigonometry Problem | AIME I, 2015 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Trigonometry.

Trigonometry Problem - AIME 2015

With all angles measured in degrees, the product (\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n\), where (m) and (n) are integers greater than 1. Find (m+n).

is 107
is 91
is 840
cannot be determined from the given information

Key Concepts

Trigonometry

Sequence

Algebra

Check the Answer

Answer: is 91.

AIME, 2015, Question 13.

Plane Trigonometry by Loney .

Try with Hints

Let (x = \cos 1^\circ + i \sin 1^\circ). Then from the identity[\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},]we deduce that (taking absolute values and noticing (|x| = 1))[|2\sin 1| = |x^2 - 1|.]

But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ - z)$, if we let our product be $M$ then[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ][= \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|]because $\sin$ is positive in the first and second quadrants.

Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1$, and so[\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.]It is easy to see that $M = 2^{89}$ and that our answer is $2+89=91$.

Sequence and Series | HANOI 2018

Try this beautiful problem from HANOI 2018 based on Sequence and Series.

Sequence and Series - HANOI 2018

Let {$u_n$} $n\geq1$ be given sequence satisfying the conditions $u_1=0$, $u_2=1$, $u_{n+1}=u_{n-1}+2n-1$ for $n\geq2$. find $u_{100}+u_{101}$

is 13000
is 10000
is 840
cannot be determined from the given information

Key Concepts

Sequence

Series

Number Theory

Check the Answer

Answer: is 10000.

HANOI, 2018

Principles of Mathematical Analysis by Rudin

Try with Hints

Here $u_2=1$, $u_3=3$, $u_4=6$, $u_5=10$

by induction $u_n=\frac{n(n-1)}{2}$ for every $n\geq1$

Then $u_n+u_{n+1}$=$\frac{n(n-1)}{2}$+$\frac{n(n+1)}{2}$=$n^{2}$ for every $n\geq1$ Then $u_{100}+u_{101}$=10000.

Number Theory and Geometry | PRMO 2019 | Problem 6

Try this beautiful problem from Pre RMO 2019 based on Number Theory and Geometry.

Number Theory and Geometry - PRMO 2019

Let abc be a three digit number with nonzero digits such that $a^{2}+b^{2}=c^{2}$. Find is the largest possible prime factor of abc.

is 13
is 29
is 840
cannot be determined from the given information

Key Concepts

Sequence

Geometry

Number Theory

Check the Answer

Answer: is 29.

PRMO, 2019

Elementary Number Theory by David Burton

Try with Hints

Here a,b,c form Pythagoras triplet then abc=345 or 435

345=(3)(5)(23) and 435=(5)(3)(29)

Then largest possible prime factor=29

Number Theory | PRMO 2019 | Problem 3

Try this beautiful problem from Pre RMO, 2019 based on the Number theory.

Number Theory - PRMO 2019

Let $x_1$ be a positive real number and for every integer $n\geq1$ let $x_{n+1}=1+x_{1}x_{2}...x_{n-1}x_{n}$. If $x_{5}=43$. what is the sum of digits of the largest prime factor of $x_{6}$.

is 13
is 25
is 840
cannot be determined from the given information

Key Concepts

Sequence

Series

Number System

Check the Answer

Answer: is 13.

PRMO, 2019

Elementary Number Theory by David Burton

Try with Hints

Here $x_5=1+x_1x_2x_3x_4$ then $x_1x_2x_3x_4=42$

$x_6=1+x_1x_2x_3x_4x_5$=1+(42)(43)=1807=(13)(139)

Then largest prime factor=139 then sum of digits=13

Limit of a Sequence | IIT JAM 2018 | Problem 2

Try this beautiful problem from IIT JAM 2018 which requires knowledge of Real Analysis (Limit of a Sequence).

Limit of a Sequence - IIT JAM 2018 (Problem 2)

Let $a_n=\frac{b_{n+1}}{b_n}$ where $b_1=1, b_2=1$ and $b_{n+2}=b_n+b_{n+1}$ , Then $\lim\limits_{n \to \infty} a_n$ is

$\frac{1-\sqrt5}{2}$
$\frac{1+\sqrt5}{2}$
$\frac{1+\sqrt3}{2}$
$\frac{1-\sqrt3}{2}$

Key Concepts

Real Analysis

Sequence of Reals

Limit of a Sequence

Check the Answer

Answer: $\frac{1+\sqrt5}{2}$

IIT JAM 2018 (Problem 2)

Advanced Calculus by Patrick Fitzpatrick

Try with Hints

Given that, $a_n=\frac{b_{n+1}}{b_n}$

$\Rightarrow \lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} \frac{b_{n+1}}{b_n}= \mathcal{L} $ (say)

Now we know that , $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} a_{n+1} $

$\Rightarrow \mathcal{L}=\lim\limits_{n \to \infty} a_{n+1}$

Can you find an equation on $\mathcal{L}$ from which the value of $\mathcal{L}$ can be obtained.

$\mathcal{L}= \lim\limits_{n \to \infty } a_{n+1}$

$= \lim\limits_{n \to \infty} \frac{b_{n+2}}{b_{n+2}}$

$=\lim\limits_{n\to \infty} \frac{b_{n+1}+b_n}{b_{n+1}}$ [By the given recurrence relation]

$=\lim\limits_{n\to \infty} \left(1+\frac{b_n}{b_{n+1}}\right)$

$=1+\lim\limits_{n \to \infty} \frac{b_n}{b_{n+1}}$

$=1+\frac{1}{\lim\limits_{n\to\infty}\frac{b_{n+1}}{b_n}}$

$=1+\frac{1}{\mathcal{L}}$

Now the value of $\mathcal{L}$ can be easily obtained

i.e., $\mathcal{L}=1+\frac{1}{\mathcal{L}}$

$\Rightarrow \mathcal{L}^2-\mathcal{L}-1=0$

$\Rightarrow \mathcal{L}=\frac{1\pm \sqrt{5}}{2}$

$\Rightarrow \mathcal{L}=\frac{1+\sqrt{5}}{2}$ [Since $a_n>0$] [ANS]

Sequence and Series: IIT JAM 2016 Problem 24

Sequence and series of real numbers

A sequence of real numbers is an one to one mapping from $\mathbb{N}$ (the set of natural numbers) to a subset of $\mathbb{R}$(the set of all real numbers).

Sum of the terms of a real sequence sequence is called a series.

Try the problem

IIT JAM 2016, PROBLEM 24

Find the sum of the series $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2+n-2}$.

$\textbf{(A)}\quad \frac13\ln 2-\frac{5}{18}\quad \textbf{(B)}\quad \frac13\ln 2-\frac{5}{6}\quad \textbf{(C)}\quad \frac23\ln 2-\frac{5}{18}\quad \textbf{(D)}\quad \frac23\ln 2-\frac{5}{6}\quad$

Knowledge Graph

Use some hints

This problem is a very basic one, this problem can easily be solved by step by step solution. The steps are:

Step 1 : First we will ignore the summation part. We will factorize the denominator, because we are going step by step so our aim is to simplify the given problem first.

Step 2: After factorizing the the denominator we will reach to a position where we have to use partial fraction to go forward.

Step 3: In this step we will take care of the $(-1)^n$ part, like how it will affect the series.

Step 4: After taking care of the $(-1)^n$ we will now expand the summation (breaking it into infinite sum).

Step 5 : So after $4$ steps we are halfway done now just the last simplification is left we will use the value

$\ln 2= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$

to simplify it further.

Now try to solve the entire problem by following these steps !!!

Let us see how to execute STEP 1 .

$\displaystyle\sum_{n=2}^{\infty} \frac{ (-1)^n }{ n^2+n-2 }$, here we are only concerned with the denominator part so,

$\frac{1}{n^2+n-2}=\frac{1}{(n+2)(n-1)}$

So we have successfully factorized the denominator. Now you can reach to the part where we have to use the concept of partial fraction!!!

Let us execute STEP 2 now.

After factorizing the denominator in HINT 1 we get

$\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{(n+2)(n-1)}$

Now our aim here is to seperate $(n+2)$ and $(n-1)$ so we can do this by using partial fraction.

$\frac{1}{(n+2)(n-1)}=\frac{A}{n+2}+\frac{B}{(n-1)}$

$\Rightarrow A(n-1)+B(n+2)=1$

Now taking $\underline{n=1}$ we get

$A(1-1)+B(1+2)=1$

$B=\frac{1}{3}$

Again taking $\underline{n=-2}$, we get:

$A(-2-1)+B(-2+2)=1$

therefore, $A=-\frac{1}{3}$

So we get, $\frac{1}{(n+2)(n-1)}=\frac{1}{3}[\frac{1}{n-1}-\frac{1}{n+2}]$

So our aim here is successful we have separated $(n+2)$ and $(n-1)$ . Now can you proceed further ???

Now after using partial fraction we get.

$\displaystyle\sum_{n=2}^{\infty} (-1)^n\frac{1}{3}[\frac{1}{n-1}+\frac{1}{n+2}]$

Now we will execute STEP 4 i.e., we will take care of the $(-1)^n$ and at the same time we will execute STEP 4 i.e, we will split it into the infinite sum.

$\frac{1}{3}\displaystyle\sum_{n=2}^{\infty} [\frac{1}{n-1}+\frac{1}{n+2}](-1)^n$

$\Rightarrow \frac{1}{3}[(1-\frac{1}{4})-(\frac{1}{2}-\frac{1}{5})+(\frac13-\frac16)\ldots]$

$\Rightarrow \frac13[(1-\frac12+\frac13-\frac14\ldots)+(-\frac14+\frac15-\frac16\ldots)]$

Now after this can you applying the formula of '$\ln 2$' to finish the problem!!

Now only STEP 5 is left to execute. So we will use the infinite series of $\ln 2$.

$\frac13[(1-\frac12+\frac13-\frac14\ldots)+(-\frac14+\frac15-\frac16\ldots)]$

$=\frac13[\ln 2+\ln 2-(1-\frac12+\frac13)]$

$=\frac13(2\ln 2-\frac56)$

$=\frac23 \ln 2 - \frac{5}{18}$

Hence the anser is C.

Books

Introduction to Real Analysis : Robert G. Bartle & Donald R. Sherbert.