rove by induction that for all n ∈ N,5 + 7 + . . . + (2n + 3)= ∑i=ni=1 (2i + 3) = n(n + 4)

rove by induction that for all n ∈ N,5 + 7 + . . . + (2n + 3)= ∑i=ni=1 (2i + 3) = n(n + 4)

If x(n) = [1,2,0,3, -2,4,7,5] evaluate the following i) X (0) ii) X (4) iii) ∑ 𝑋(𝐾) 7 𝑘=0 iv) ∑ |𝑋(𝐾)| 7 𝑘=0 2

Prove the following statement by mathematical induction.For every integer n ≥ 0, 7n − 2n is divisible by 5.Proof (by mathematical induction): Let P(n) be the following sentence.7n − 2n is divisible by 5.We will show that P(n) is true for every integer n ≥ 0.Show that P(0) is true: Select P(0) from the choices below.5 | (70 − 20)70 − 20 < 5    (70 − 20) | 55 is a multiple of 70 − 20The truth of the selected statement follows from the definition of divisibility and the fact that 70 − = 0.Show that for each integer k ≥ 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k ≥ 0, and suppose that P(k) is true. Select P(k) from the choices below.5 is a multiple of 7k − 2k5 is divisible by 7k − 2k    7k − 2k is divisible by 57k − 2k < 5[This is P(k), the inductive hypothesis.]We must show that P(k + 1) is true. Select P(k + 1) from the choices below.5 is a multiple of 7k + 1 − 2k + 17k + 1 − 2k + 1 < 5    7k + 1 − 2k + 1 is divisible by 55 is divisible by 7k + 1 − 2k + 1

6n 2 +7n−11=5n 2

If n is an even integer, then (i)n is equal to