Mathematical induction!

I need this exercises
show that... is true

number 1:
1^4+2^4+3^4+...+n^4= n(n+1)(6n^3+9n^2+n-1)/30

number 2
(1+a)^n=(or bigger) 1+an

please I need this for the school! (I'L GIVE YOU POINTS)

In order to show that these statements are true by Mathematical Induction, we need to show that it holds for the smallest integer that the statement is said to be true for (in this case, the first statement is true for integers n ≥ 1, so we need to show that it holds for n = 1), and then show that if it holds for some integer k, then it is true for k + 1.

(1) First, we need to show that this holds for n = 1. For n = 1, the statement reduces to:
1^4 = 1(1 + 1)(6 + 9 + 1 - 1)/30,

which is, indeed, true. Now, if this statement is true for some integer k, then we have:
( ** ) 1^4 + 2^4 + 3^4 + ... + k^4 = k(k + 1)(6k^3 + 9k^2 + k - 1)/30.

Now, remember, we are only assuming that this is true, but it doesn't have to be true. How induction works is that we prove that it is true for a specific integer, and then showing that if it works for k + 1 if it works for some integer k allows us to show that this is true. For example, here we have shown that it works for n = 1. If we can show that if this works for k + 1 if it also works for some integer k, we can use this to say that this statement also works for n = 1 + 1 = 2, which can be used again to show that this also works for n = 2 + 1 = 3, and so on.

We now need to show that:
1^4 + 2^4 + 3^4 + ... + k^4 + (k + 1)^4 = (k + 1)[(k + 1) + 1][6(k + 1)^3 + 9(k + 1)^2 + (k + 1) - 1]/30
==> 1^4 + 2^4 + 3^4 + ... + k^4 + (k + 1)^4 = (k + 1)(k + 2)[6(k + 1)^3 + 9(k + 1)^2 + k]/30.

In order to do this, note that:
1^4 + 2^4 + 3^4 + ... + k^4 + (k + 1)^4 = (1^4 + 2^4 + 3^4 + ... + k^4) + (k + 1)^4
= k(k + 1)(6k^3 + 9k^2 + k - 1)/30 + (k + 1)^4, from ( ** )
= [k(k + 1)(6k^3 + 9k^2 + k - 1) + 30(k + 1)^4]/30