The sum of the cubes of three consecutive positive integers

Question

1. Prove by induction that it is possible to pay, without requiring change, any whole number of roubles greater than 7 with banknotes of value 3 roubles and 5 roubles.

2. Prove by induction that _r=1Σⁿ r² = (1/6) (n +1) (2n + 1).

Deduce formulae for

1·1 +2·3 + 3·5 + 4·7 + ········ + n (2n- 1) and 1² + 3² + 5²+ · · · · ·  (2n-1)².

3. Here is another way to work out _r=1Σⁿ r². Observe that (r+ 1)³ - r³ = 3r² + 3r + 1. Hence

_r=1Σⁿ r²(r+ 1)³ - r³ = 3 _r=1Σⁿ r² +3 _r=1Σⁿ r + n

The left-hand side is equal to

(2³ - 1³) + (3³ -2³) + (4³ -3³) +······ + ((n +1)³ -n³) = (n +1)³ -1. Hence we can work out _r=1Σⁿ r².

Carry out this calculation, and check that your formula agrees with that in Exercise 2.

Use the same method to work out formulae for _r=1Σⁿ r³ and _r=1Σⁿ r⁴.

4. Prove the following statements by induction:

(a) For all integers n ≥ 0, the number 5²ⁿ- 3ⁿ is a multiple of 11.

(b) For any integer n ≥ 1, the integer 2⁴ⁿ-1 ends with an 8.

(c) The sum of the cubes of three consecutive positive integers is always a multiple of 9.

(d) If x ≥ 2 is a real number and n ≥1I is an integer, then xⁿ ≥ nx.

(e) If n > 3 is an integer, then 5ⁿ > 4ⁿ + 3ⁿ + 2ⁿ.

5. a) Factorize x^{2n +1} - 1 as a product of real linear and quadratic polynomials.

(b) Write x²ⁿ + x^2n-1 + ······ x + 1 as a product of real quadratic polynomials.

(c) Let ω = e ^{(2Πi) / (2n +1)+1}. Show that Σ ω^i+j = 0, where the sum is over all I and j from1 to 2n + 1 such that i < j.

6. Here is a "proof" by induction that any two positive integers are equal (e.g., 5 = 10):

First, a definition: if a and b are positive integers, define max (a, b) to be the larger of a and b if a ≠ b, and to be a if a = b. [For instance, max (3, 5) = 5, max (3, 3) = 3.] Let P (n) be the statement: "if a and b are positive integers such that max (a, b) = n, then a = b." We prove P (n) true for all n > 1 by induction. [As a consequence, if a, b are any two positive integers, then a = b, since P (n) is true, where it = max (a, b).] First, P (1) is true, since if max (a, b) = 1 then a and b must both be equal to I. Now assume P (n) true. Let a, b be positive integers such that max (a, b) = n + 1. Then max (a - 1, b - 1) = n. As we are assuming P (n), this implies that a - 1 = b - 1, hence a = b. Therefore, P (n + I) is true. By induction, P (n) is true for all n.

There must be something wrong with this "proof" Can you find the error?