Please help, Mathematical Induction?

Question

Use mathematical induction to prove that:

(a) 2^2 + 4^2 +…+ (2n)^2 = [2n (n+1) (2n+1)]/3, for n ≥ 1, and

(b) 5n! ≥ 3^n, for n ≥ 2

Need full explanation of steps for revision.

Answer 1

(A) Show base case is true:

(2*1)^2 = (2*1)(2)(2*1 + 1) / 3
4 = 4

Assume the equality holds for n = k
Show that it holds for n = k+1

2^2 + 4^2 + 6^2 + ... + (2k)^2 + (2(k+1))^2
= 2k (k+1) (2k + 1) / 3 + (2k+2)^2
= 2k (k+1) (2k + 1) / 3 + (4k^2 + 8k + 4)
= (2k^2 + 2k) (2k + 1) / 3 + (12k^2 + 24k + 12) / 3
= (4k^3 + 6k^2 + 2k + 12k^2 + 24k + 12) / 3
= (4k^3 + 18k^2 + 26 k + 12) / 3

2(k+1)(k+2)(2k+3) / 3
= 2 (k^2 + 3k + 2)(2k + 3) / 3
= 2 (2k^3 + 3k^2 + 6k^2 + 9k + 4k + 6) / 3
= (4k^3 + 18k^2 + 26k + 12) / 3

(B) Show base case is true
5 (2!) ≥ 3^2
10 ≥ 9

Assume it is true for all n < k, Show it is true for n = k+1

5(k+1)! ≥ 5 (k+1) k! ≥ 5 * 3 * 3^k ≥ 5 * 3^(k+1)

Answer 2

For b, check for n=2. 10>9. check.

Assume it holds for n, and show it for n+1.

5(n+1)!>3^(n+1)? =
5n!(n+1)>(3^n)*3.

All you need to do is check that n+1 is greater than or equal to 3, and you're done.

For a, same sort of thing. Check it for 1, and then assume it works for n, and show for n+1.

On the left, you get E(2n)^2+(2(n+1))^2, where E represents the sum of the first n terms. On the right, you get 2(n+1)(n+2)(2n+3)/3. Just expand it, check it, and you're done.