 Prove that for any we have
 For and prove that
 Again for prove

Proof. First we rewrite the sum on the right in summation notation, and then use the distributive property to compute,
This is the identity requested

Proof. From the Binomial theorem we have
Thus,
since the second term is strictly positive for positive integers. This establishes the inequality on the left.
For the inequality on the right we use part (a) to get,Thus,
In the second to last line we have just replaced each term in the sum by , giving the inequality. This establishes both sides of the requested inequality

Proof. The proof is by induction. For the case we have,
For a positive integer , then ; hence, the inequalities hold in the case . Assume then that they hold for some .
For the left inequality,This establishes the left inequality for all .
For the right inequality, assume it is true for some , thenHence, the right inequality is true for all . Therefore, the inequalities requested are indeed true for all
In b) in the second line after “thus”:
[n^{p+1}(p+1)+n^{p} + \sum_{k=0}^{p1}{\binom{p+1}{k} n^k}]
On the 4th line of your first proof, for the second summation it should be b^(pk1) instead of b^(pk)
It should be b^(pk+1)