Maths Olympiad Prep

Track / Stage 8 / 22 of 180 #1722 of 1964

Problem 1722

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.

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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
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Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.