Maths Olympiad Prep

Track / Stage 8 / 21 of 180 #1721 of 1964

Problem 1721

IMO Shortlist mid-range; USAMO P2/P5
Number theory Difficulty 8.0 Prove it

Let Tk=k−1T_k = k - 1 for k=1,2,3,4k = 1, 2, 3,4 and
T2k−1=T2k−2+2k−2,T2k=T2k−5+2k(k≥3).T_{2k-1} = T_{2k-2} + 2^{k-2}, T_{2k} = T_{2k-5} + 2^k \qquad (k \geq 3).
Show that for all kk,
1+T2n−1=[1272n−1]and1+T2n=[1772n−1],1 + T_{2n-1} = \left[ \frac{12}{7}2^{n-1} \right] \quad \text{and} \quad 1 + T_{2n} = \left[ \frac{17}{7}2^{n-1} \right],
where [x][x] denotes the greatest integer not exceeding x.x.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We will use mathematical induction to prove the given statements. Let's start by verifying the base cases and then proceed with the induction steps.

### Base Cases:
For n=1 n = 1 :
T1=1−1=0 T_1 = 1 - 1 = 0
T2=2−1=1 T_2 = 2 - 1 = 1
1+T1=1 1 + T_1 = 1
1+T2=2 1 + T_2 = 2

We need to check:
1+T2⋅1−1=⌊127⋅21−1⌋=⌊127⌋=1 1 + T_{2 \cdot 1 - 1} = \left\lfloor \frac{12}{7} \cdot 2^{1-1} \right\rfloor = \left\lfloor \frac{12}{7} \right\rfloor = 1
1+T2⋅1=⌊177⋅21−1⌋=⌊177⌋=2 1 + T_{2 \cdot 1} = \left\lfloor \frac{17}{7} \cdot 2^{1-1} \right\rfloor = \left\lfloor \frac{17}{7} \right\rfloor = 2

Both base cases hold true.

### Induction Hypothesis:
Assume that for some n≥1 n \geq 1 , the following statements hold:
1+T2n−1=⌊127⋅2n−1⌋ 1 + T_{2n-1} = \left\lfloor \frac{12}{7} \cdot 2^{n-1} \right\rfloor
1+T2n=⌊177⋅2n−1⌋ 1 + T_{2n} = \left\lfloor \frac{17}{7} \cdot 2^{n-1} \right\rfloor

### Induction Step:
We need to show that:
1+T2(n+1)−1=⌊127⋅2n⌋ 1 + T_{2(n+1)-1} = \left\lfloor \frac{12}{7} \cdot 2^n \right\rfloor
1+T2(n+1)=⌊177⋅2n⌋ 1 + T_{2(n+1)} = \left\lfloor \frac{17}{7} \cdot 2^n \right\rfloor

Using the given recurrence relations:
T2(n+1)−1=T2(n+1)−2+2(n+1)−2=T2n+2n−1 T_{2(n+1)-1} = T_{2(n+1)-2} + 2^{(n+1)-2} = T_{2n} + 2^{n-1}
T2(n+1)=T2(n+1)−5+2n+1=T2n−3+2n+1 T_{2(n+1)} = T_{2(n+1)-5} + 2^{n+1} = T_{2n-3} + 2^{n+1}

By the induction hypothesis:
1+T2n=⌊177⋅2n−1⌋ 1 + T_{2n} = \left\lfloor \frac{17}{7} \cdot 2^{n-1} \right\rfloor
1+T2n−3=⌊127⋅2n−2⌋ 1 + T_{2n-3} = \left\lfloor \frac{12}{7} \cdot 2^{n-2} \right\rfloor

Now, let's verify the induction step for T2(n+1)−1 T_{2(n+1)-1} :
1+T2(n+1)−1=1+T2n+2n−1 1 + T_{2(n+1)-1} = 1 + T_{2n} + 2^{n-1}
1+T2n=⌊177⋅2n−1⌋ 1 + T_{2n} = \left\lfloor \frac{17}{7} \cdot 2^{n-1} \right\rfloor
1+T2(n+1)−1=⌊177⋅2n−1⌋+2n−1 1 + T_{2(n+1)-1} = \left\lfloor \frac{17}{7} \cdot 2^{n-1} \right\rfloor + 2^{n-1}

Since 2n−1 2^{n-1} is an integer, we can write:
⌊177⋅2n−1⌋+2n−1=⌊177⋅2n−1+2n−1⌋ \left\lfloor \frac{17}{7} \cdot 2^{n-1} \right\rfloor + 2^{n-1} = \left\lfloor \frac{17}{7} \cdot 2^{n-1} + 2^{n-1} \right\rfloor
=⌊(177+1)⋅2n−1⌋ = \left\lfloor \left( \frac{17}{7} + 1 \right) \cdot 2^{n-1} \right\rfloor
=⌊247⋅2n−1⌋ = \left\lfloor \frac{24}{7} \cdot 2^{n-1} \right\rfloor
=⌊127⋅2n⌋ = \left\lfloor \frac{12}{7} \cdot 2^n \right\rfloor

Thus, the first part of the induction step holds.

Now, let's verify the induction step for T2(n+1) T_{2(n+1)} :
1+T2(n+1)=1+T2n−3+2n+1 1 + T_{2(n+1)} = 1 + T_{2n-3} + 2^{n+1}
1+T2n−3=⌊127⋅2n−2⌋ 1 + T_{2n-3} = \left\lfloor \frac{12}{7} \cdot 2^{n-2} \right\rfloor
1+T2(n+1)=⌊127⋅2n−2⌋+2n+1 1 + T_{2(n+1)} = \left\lfloor \frac{12}{7} \cdot 2^{n-2} \right\rfloor + 2^{n+1}

Since 2n+1 2^{n+1} is an integer, we can write:
⌊127⋅2n−2⌋+2n+1=⌊127⋅2n−2+2n+1⌋ \left\lfloor \frac{12}{7} \cdot 2^{n-2} \right\rfloor + 2^{n+1} = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊(127⋅2n−2+2n+1)⌋ = \left\lfloor \left( \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right) \right\rfloor
=⌊(127⋅2n−2+2n+1)⌋ = \left\lfloor \left( \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right) \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
=⌊127⋅2n−2+2n+1⌋ = \left\lfloor \frac{12}{7} \cdot 2^{n-2} + 2^{n+1} \right\rfloor
\[ = \left\lfloor \

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.