This is an optional module for the students preparing for engineering entrance examinations
Q-1) Show that tere cannot be an infinite AP, all of whose terms are perfect squares.
Solution:
Assuming that there exists such an AP, let \(d\) denote the common difference of this progression. Thus \(d\) can be represented as the difference of 2 perfect squares in infinitely many ways. Note that if \(x^2-y^2=d\) for \(x,y\in \mathbb{Z}\), we have \(|x+y| \le |d|\) and \(|x-y|\le|d|\) which basically implies that the maximum value of \(|x|\) is limited to \(|d|\). This contradicts the assumption that the equation \(d=x^2-y^2\) holds true for infinitely many integers \(x,y\). Thus no such AP exists.
Q-2) Let \(x_1\) be a given positive integer. A sequence \(\left\{x_n\right\}_{n\ge 1}\) of positive integers is such that \(x_n\), for \(n\ge 2\), is obtained from \(x_{n-1}\) by adding some nonzero digit of \(x_{n-1}\). Prove that
a) the sequence contains an even term;
b) the sequence contains infinitely many even terms.
Solution:
We note that the given sequence is strictly increasing. Also, \(x_n<x_{n+1}<x_n+10\), this implies that the set of numbers between any consecutive multiples of \(10\) greater than \(x_1\) must contain a term of this sequence.
Let \(m\) denote the number of digits of \(x_1\). Consider \(x_k \in [10^m,10^m+1]\). If \(x_k\) is even, we are done with the first proof, else note that the only non-zero digits of \(x_k\) are the first digit (which is \(1\)) and the last digit which is odd. This implies that \(x_{k+1}\) is even. Note that this holds for any \(m\) which proves the second point.
Q-3) Define a sequence \(\left\{u_n\right\}\) by \(u_0=u_1=u_2=1\) and thereafter by the condition \(u_nu_{n+3}-u_{n+1}u_{n+2}=n!\). We have already defined that \(n!=n\times (n-1)\times (n-2)\times….2\times 1\). Also \(0!=1\). Show that \(u_n\) is an integer for all \(n\).
Solution:
From the recurrence relation we see that
Similarly we have \(u_5=8\), \(u_6=15=5\times 3\), \(u_7=48=6\times 4\times 2\), \(u_8=105=7\times 5\times 3\), \(u_9=384=8\times 6\times 4\times 2\).
We inductively claim that \(u_n=(n-1)(n-3)(n-5)….3\)(or \(2\)).
Assume that this claim holds for all \(k\le n+2\), thus
Q-4) Let \(Q_0(x)=1\), \(Q_1(x)=x\) and define, for \(n\ge 2\),
Show that \(Q_n(x)\) is a polynomial with integer coefficients for all integers \(n\).
Solution:
We have from the given recurrence relation:
From the last 2 equations we have
Since \(Q_k(x)\) isn’t the identity/zero polynomial for any \(k\) (in fact, we can inductively show that the degree of \(Q_n(x)\) is \(n\)), we can have:
This means that the fraction \(\frac{Q_n(x)+Q_{n-2}(x)}{Q_{n-1}(x)}\) is a constant. We know that \(Q_2(x)=x^2-1\), thus putting \(n=2\), we find this ratio to be \(x\).
We thus have the recurrence:
\(Q_n(x)=xQ_{n-1}(x)-Q_{n-2}(x)\), and induction proves that \(Q_n(x)\) is a polynomial with integer coefficients for all \(n\in \mathbb{N}\).
Q-5) Let \(a_0=1\), \(a_1=2\) and for \(n \ge 2\), define \(a_n=4a_{n-1}-a_{n-2}\).
a) Show that \(a_n\) divides \(a_{kn}\) for all positive integers \(n\) and odd positive integer \(k\).
b) Using (a) or otherwise, find an odd prime factor for \(a_{2015}\).
Solution:
From the method of difference equations, we have :
Note that \(a_n \in \mathbb{Z}\) for all naturals \(n\).
where \(\alpha,\beta\) are the roots of the characteristic difference equation of the given sequence. Since \(k\) is odd, and \(\alpha, \beta\) are conjugate rationals, we conclude that the ex[ression on the RHS is an integer.
Putting \(n=5,k=43\), we conclude that \(a_{2015}\) is divisible by \(a_5=362\), thus \(181\) is an odd prime factor of \(a_{2015}\).
Q-6) Define a sequence \(a_n\) for \(n \ge 1\) by \(a_1=1\) and \(a_2=2\) and \(a_{n+2}=2a_{n+1}-a_n+2\) for \(n>2\). prove that for any \(m\), \(a_ma_{m+1}\) is also a term in this sequence.
Solution:
We have \(a_1=1\), \(a_2=2\), \(a_3=5\), \(a_4=10\).
Inductively, if we claim that for all integers till \(n+1\), \(a_n=(n-1)^2+1\), then we have
\(a_{n+2}=2n^2+2-(n^2-2n+2)+2=n^2+2n+2=(n+1)^2+1\).
Thus we have:
\(a_ma_{m+1}=[(m-1)^2+1][m^2+1]=[m(m-1)]^2+2m^2-2m+2=[m(m-1)+1]^2+1=a_{m(m-1)+2}\).
Q-7) For a sequence \(\left\{ a_n\right\}\), we have \(a_0=a_n=0\) and \(a_{k-1}-2a_k+a_{k+1}\ge 0\) for \(k=1,2,…n-1\). Show that \(a_k \ge 0\) for all \(k\).
Solution:
From the given recurrence, we can conclude that \(a_{k+1}-a_k \ge a_k-a_{k-1}\) for all \(k \in [1,n-1]\). If we graphically visualise this, we see that the broken line segments with vertices \((i,a_i)\) is thus convex, and the slope of each succeeding segment is greater than or equal to that of its preceeding segment.
If for some \(p\), we have \(a_{p-1} \le 0\) and \(a_p>0\) then we can conclude that
\(a_n-a_{n-1}\ge a_{n-1}-a_{n-2}\ge….\ge a_p-a_{p-1}>0 \). This implies
\(a_n>a_{n-1}>…>a_p>0\) contradicting the given.
Q-8) Define a sequence by \(a_1=0\) and \(|a_i|=|a_{i+1}+1|\) for \(i>1\).
Show that \(\frac{a_1+a_2+a_3+…a_n}{n}\ge -\frac{1}{2}\).
Solution:
We have \(\sum\limits_{i=1}^{n+1}a_i^2=\sum\limits_{i=2}^n(a_i+1)^2\).
This implies \(a_{n+1}^2=2(a_1+a_2+..a_n)+n \ge 0\Rightarrow \frac{a_1+a_2+a_3+…a_n}{n}\ge -\frac{1}{2}\)
Q-9) Show that there cannot exist a sequence \(\left\{ a_n\right\}\) of non-negative integers such that \(a_{mn}=a_m+a_n\) for all \(m,n \in \mathbb{N}\).
Solution:
Putting \(n=1\), we obtain \(a_1=0\), thus \(a_k \ge k-1\) for all natural numbers \(k\).
Put \(m=2\) to have
\(a_{2n}=a_n+a_2\), and also for \(m=4\), we have \(a_{4n}=a_4+a_n=2a_2+a_n\).
For \(m=2^k\) we have \(a_{2^kn}=ka_2+a_n\), now putting \(n=1\) we get
\(a_{2^k}=ka_2 \ge 2^k-1\Rightarrow a_2 \ge \frac{2^k-1}{k}\) for all natural numbers \(k\) which is impossible. Thus no such sequence exists.
Q-10) A sequence is defined by \(a_1=1\), and \(a_{n+1}=a_n+\frac{1}{a_n^2}\).
a)Show that \(\left\{a_n\right\}\) is unbounded.
b) Show that \(a_{9000}>30\).
Solution:
Cubing both sides of the recurrence relation gives:
\(a_{n+1}^3=a_n^3+\frac{1}{a_n^6}+\frac{3}{a_n}\left(a_n+\frac{1}{a_n^2}\right)>a_n^3+3\).
Also \(a_2=2\Rightarrow a_2^3=8>3\times 2=6\). So inductively we claim that \(a_n^3>3n\Rightarrow a_{n+1}^3>a_n^3+3=3n+3\Rightarrow a_{n+1}^3>3(n+1)\). Our induction ends over here.
Thus we see that \(a_n>\sqrt[3]{3n}\), RHS is an unbounded sequence and thus we show that \(\left\{a_n\right\}\) is an unbounded sequence.
We have \(a_n>\sqrt[3]{3n}\Rightarrow a_{9000}>\sqrt[3]{27000}=30\).
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