Question 836817
a) Base cases n = 1 and n = 2 holds (you will see why I checked two cases). Assume that for some *[tex \large n \ge 2], that *[tex  \large x-y | x^n - y^n] and *[tex \large x-y| x^{n-1} - y^{n-1}]. We look at the n+1th term:


*[tex \large x^{n+1} - y^{n-1} = (x^n - y^n)(x+y) - x^ny + y^nx = (x^n - y^n)(x+y) - xy(x^{n-1} - y^{n-1})]


The RHS is divisible by x-y by our induction hypothesis, so the LHS must also be divisible by x-y, for all natural numbers n.


b) Again, base case n = 1 holds. Assume that the statement holds for some *[tex \large n \ge 1]. Then, by our hypothesis,


*[tex \large \sum_{i=1}^{n+1} i \times i! = \left( \sum_{i=1}^{n} i \times i! \right) + (n+1) \times (n+1)!]

*[tex \large = (n+1)! - 1 + (n+1) \times (n+1)!]

*[tex \large = (n+2) \times (n+1)! - 1]

*[tex \large = (n+2)! - 1]


Therefore the statement holds for n+1, so by induction, we are done.