Discrete Math. Practice. For students of technical specialties. Ivan Treschev
for n = 1 we have: 2> 1.
Induction step: let the inequality be valid for k = n, we prove the inequality for k = n +1: 2n+1> n+1
2*2n> n+1 (1) We
use the property: if a> b and b ≥ c, then a> c, i.e. we replace in (1) 2n with n, we have: 2n ≥ n+1
n ≥ 1 (2)
Studying inequality (2), we can conclude that 2n> n only for n ≥ 1, and since. n ∈ N, then this condition is satisfied, the inequality is proved.
Problem number 10: prove the equality: 12+22+32+…+n2= ((1) \ (6)) n (n+1) (2n+1).
Solution: we prove the inequality using the mathematical induction method: Induction
base: for n = 1 we have: 1= ((1) \ (6)) *2*3=1
Induction step: let the inequality be valid for k = n, we prove the inequality for k = n +1: 12+22+32+…+n2+ (n+1) 2= ((1) \ (6)) (n+1) (n+1+1) (2 (n+1) +1)
12+22+32+…+n2+ (n+1) 2= ((1) \ (6)) (n+1) (n+2) (2n+3) (1)
Replace in (1) {12+22+32+…+n2} by {((1) \ (6)) n (n+1) (2n+1)}, we get: ((1) \ (6)) n (n+1) (2n+1) + (n+1) 2= ((1) \ (6)) (n+1) (n+2) (2n+3)
n (2n+1) +6 (n+1) = (n+2) (2n+3)
2n2+n+6n+6 = 2n2+3n+4n+6
2n2+7n+6 = 2n2+7n+6
Equality is proved.
Problem number 11: there is a chessboard, a horse is located in the lower left corner. Cut the bottom right corner from the board. Question: is it possible to get around such a chessboard with a knight?
Solution: it is easy to notice that, making a move with the knight, we alternate the color of the cell on which it is located, i.e. if we started with a black cell our movement, then we must finish on white. It is also clear that the left and right lower cells are not the same color. So at the last step we have 2 cells left (a horse is standing on one of them), which we have not yet walked around, and they are of the same color.
Therefore, it is impossible to get around such a chessboard with a horse.
Problem number 12: to prove the inequality: ((1) \ (√ 1)) + ((1) \ (√ 2)) + ((1) \ (√ 3)) + ((1) \ (√ 4)) +…+ ((1) \ (√ n)) + ≥ √ n for n≥0.
Solution: we prove the inequality using the method of mathematical induction:
Base of induction: for n = 1 we have:1=1.
Induction step: suppose that the inequality is true for k = n, we prove the inequality for k = n +1:
((1) \ (√ 1)) + ((1) \ (√ 2)) + ((1) \ (√ 3)) + ((1) \ (√ 4)) +…+ ((1) \ (√ n)) + ((1) \ (√ n+1)) ≥ √ n+1 (1) We
use the property: if a> b and b ≥ c, then a> c i.e. replace in (1) (((1) \ (√ 1)) + ((1) \ (√ 2)) + ((1) \ (√ 3)) + ((1) \ (√ 4)) +…+ ((1) \ (√ n))) by √ n, we have: √ n + ((1) \ (√ n+1)) ≥ √ n+1
√ n ≥ √ n+1 – ((1) \ (√ n+1))
√ n ≥ ((n+1—1) \ (√ n+1))
n ≥ ((n2) \ (n+1))
n2+n ≥ n2
n ≥ 0 (2)
Studying inequality (2), we can conclude that ((1) \ (√ 1)) + ((1) \ (√ 2)) + ((1) \ (√ 3)) + ((1) \ (√ 4)) +…+ ((1) \ (√ n)) + ≥ √ n, only for n ≥ 0, which coincides with the initial conditions, the inequality is proved.
Task number 13:share the booty n robbers. Each of them has their own opinion on the value of a particular share of production, and each of them wants to get no less than (1 /n) the share of production (from their point of view). How to divide prey between robbers?
Solution: for two robbers, the task is not difficult to solve – one divides the production into two equal shares in his opinion, and the other selects the largest share from them. We will solve the problem by induction on the number of robbers, i.e. Suppose k robbers already have a way to split prey harmlessly. We will divide prey between (k +1) robbers. We divide all the prey between k robbers and then let each of them divide his share into (k +1) equal parts in his opinion. Now let the last robber select one of these parts from each of the k robbers. The last robber took (in his opinion) no less than [1 / (k +1)] the share of each of the k robbers, i.e. In total, he received at least [1 / (k +1)] from all production. Each of the first k robbers also received at least (1 / (k)) * (k / (k+1)) = (1 / (k+1)) from all production.
Problem number 14: prove that an equilateral triangle cannot be covered by two smaller equilateral triangles.
Proof: Each of the smaller triangles cannot cover more than one vertex of a large triangle. According to the Dirichlet principle, there are more cells (in this case, 3 vertices) than rabbits (2 vertices). Therefore, an equilateral triangle cannot be covered by two smaller equilateral triangles.
Problem number 15: prove that (1+x) n ≥ 1+nx.
Proof: (1+x) n+1 ≥ 1+ (n+1) x
(1+x) n (1+x) ≥ (1+nx) +x
(1+x) (1+x) n ≥ (1+x) (1+nx) +x
(1+x) (1+nx) =1+nx+x+nx2
1+nx+x ≤ 1+nx+x+nx2
That is: (1+x) (1+x) n ≥ (1+x) (1+nx) ≥ (1+nx) +x
Problem 16 if (x+ ((1) \ (x))) – an integer, whether integer xn+ ((1) \ (x)) n?
Proof: welimit the answer by induction. For n = 0: 1 +1 = 2. If n = -m, where m is the positive integer: xn+ ((1) \ (xn)) = x-m+ ((1) \ (x-m)) = xm+ ((1) \ (xm))
Let xk+ ((1) \ (xk)) are integers (k=0,…,k). Let us prove that xk+1+ ((1) \ (xk=1)) is also an integer: [(xk+ ((1) \ (xk))) (x+ ((1) \ (x)))] = xk+1+ ((1) \ (xk-1)) + xk-1+ ((1) \ (xk+1)) = [(xk+1+ ((1) \ (xk+1))) + (xk-1+ ((1) \ (xk-1)))] ⇒ xk+1+ ((1) \ (xk+1)) = (xk+ ((1) \ (xk))) (x+ ((1) \ (x))) – (xk-1+ ((1) \ (xk-1)))
The first term on the right-hand side is the product of integers by the assumption of induction, therefore, the integer, the second is similar. So, the sum is an integer, therefore, the left side is an integer.
Problem number 17: qprove that a number made up of 3n identical digits is divisible by 3n.
Proof: Fix one of the numbers – a. Denote the number made up of 3nidentical digits aby An. We use induction on n. Obviously, A1 is divisible by 3. Further, suppose that Akis divisible by 3k. Note that Ak +1 = Ak * 100 … 0100 … 01 (the number of zeros in each ellipse is 3k – 1). Since the sum of the digits of the number 100 … 0100 … 01 is 3, it is divided by 3, and therefore, Ak +1is divided by 3k * 3 = 3k +1. By induction, the statement of the problem is proved.
Problem number 18: how many ways can decompose the number 1024 into a product of three natural numbers, each of which is greater than 1.
Solution: This is the number of solutions to the equation x1+ x2,