= (k + 1)(6k^4 + 39k^3 + 91k^2 + 89k + 30)/30.
At this point, comparing this to ( ** ) reveals that we need to show that:
6k^4 + 39k^3 + 91k^2 + 89k + 30 = (k + 2)[6(k + 1)^3 + 9(k + 1)^2 + k],
which you can do by just expanding the right side. Since this above equality is true, we can deduce that:
(k + 1)(6k^4 + 39k^3 + 91k^2 + 89k + 30)/30 = (k + 1)(k + 2)[6(k + 1)^3 + 9(k + 1)^2 + k],
and so the statement holds for k + 1 if it holds for some integer k. Therefore, by the principle of Mathematical Induction, the statement in question is true.
(2) Assuming that we need to show that this is true for n ≥ 0, we need to show that this holds for n = 0. It does, indeed, hold when n = 0 since:
(1 + a)^0 ≥ 1 + a(0) ==> 1 ≥ 1.
Now, assuming that (1 + a)^k ≥ 1 + ak holds, we need to show that:
(1 + a)^(k + 1) ≥ 1 + a(k + 1) ==> (1 + a)^(k + 1) ≥ ak + a + 1.
To do this, note that:
(1 + a)^(k + 1) = (1 + a)(1 + a)^k
≥ (1 + a)(1 + ak), by the induction hypothesis
= 1 + ak + a + a^2*k, by FOILing
≥ 1 + ak + a + 0, since a^2*k ≥ 0 for k ≥ 0
= ak + a + 1.
Since we have shown that this holds when n = k + 1 when it holds for n = k, the proof is complete.
I hope this helps!