Mathematical induction!
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Mathematical induction!

[From: ] [author: ] [Date: 12-08-05] [Hit: ]
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.......
= (k + 1)[k(6k^3 + 9k^2 + k - 1) + 30(k + 1)^3]/30, by factoring out k + 1
= (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!

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Part one is a proof by induction.

Assume it true for n.

Then show it is true for n+1. i.e. show that n(n+1)(6n^3+9n^2+n-1)/30 + (n+1)^4 =

(n+1)(n+2)(6(n+1)^3+9(n+1)^2+(n+1)-1)/…

Then show it is true for n=1 i.e 1^4 = 1 equals 1 (2) (6+9)/30= 1 so it then must be true for 1+1, etc.

Number 2 don't understand what the question is unless it is

(1+a)^n >= 1+an

(1+a)^n = 1^n + n 1^(n-1) a + (n/2)(n-1) 1^(n-2)a^2 ......

= 1 + an + (n/2)(n-1)a^2 ........
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