Question

read the example and froof and answer for question 2.

Example: Prove Vn EZ with n20, 8 (3-1) Proof: Let P(n) be 8 (3*-1). [Again, using the word "be" since using an equals sign with a divisibility symbol would make no sense.] Since 320-1-0 and 8 0, P(o) is true. Next, let k eZ and k 20 and assume P(k) is true. This means 8|(32-1) so 3 xeZ such that 8x 3-1, or 3 8x+1. Then 32+)-13242 -1 -3 32-1 -9(8x+1)-1 Since 9x +1eZ, P(k+I) is true [because 8 (3d-)-1)]. Therefore, P(n) is true Vn e Z with n20,

Answer 1

Given data n^3 - 7n + 3 is divisible by 3 For n = 1, = n^3 - 7n + 3 = (1)^3 - 7(1) + 3 = 1 - 7 + 3 = -6 + 3 = -3 -3 is divisible by 3 For n = k it is true therefore k^3 - 7k + 3 is divisible by 3 n = k + 1, we get therefore k + 1 is true = (k + 1)^3 - 7(k + 1) + 3 = k^3 + 3k^2 + 3k + 1 - 7k - 7 + 3 = (k^3 - 7k + 3) + (3k^2 + 3k - 6) = (k^3 - 7k + 3) + 3(k^2 + k - 2) since the coefficient is 3 The 1st and 2nd parts are is divisible by 3 \r\n therefore (k + 1)^3 - 7(k + 1) + 3 is divisible by 3 p(k + 1) is true p(n) is true for all n greaterthanorequalto 0 therefore n^3 - 7n + 3