SOLUTION: Prove by induction that for all n (n being positive natural numbers),
a) (x^n) - (y^n) is a multiple of (x-y), where x does not equal y, and x and y are integers.
b) (Sigma
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Question 836817: Prove by induction that for all n (n being positive natural numbers),
a) (x^n) - (y^n) is a multiple of (x-y), where x does not equal y, and x and y are integers.
b) (Sigma n, i=1) i x i! = (n+1)! - 1
This is a discrete mathematics word problem.
Found 2 solutions by stanbon, richard1234:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Prove by induction that for all n (n being positive natural numbers),
a) (x^n) - (y^n) is a multiple of (x-y), where x does not equal y, and x and y are integers.
---
1st: Show it is true for n = 1
(x^1)-(y^1) = 1(x-y):::: true
-----
2nd: Assume it is true for n = k::
(x^k)-(y^k) = m(x-y) where m is an integer
-----------
3rd: Prove it is true for n = k+1
(x^(k+1))-(y^(k+1)) = (x^k)*x - (y^k)*y
= (x^k)-(y^k) + x-y
Then (x^k-y^k) = m(x-y) and (x-y) = 1*(x-y)
So, (x^(k+1))-(y^(k+1)) = (m+1(x-y)
QED
------------------------------------------------------
b) (Sigma n, i=1) i x i! = (n+1)! - 1
1st Show for i = 1
1x1! = (1+1)!-1
1*1 = = 2!-1 = 1
1 = 1
----------------
2nd: Assume true for i = k
k*k! = (k+1)! -1
---------------------------
3rd: Prove true for i = k+1
(k+1)(k+1)!
= k(k+1)! + (k+1)!
= k(k+1)k! + (k+1)k!
= (k+1)[k(k!] + k*k!+ k!
= (k+1+1][k*k!] + k!
From 2nd you get::
= (k+2][(k+1)!-1] + k!
= (k+2)(k+1)! - (k+2) + k!
= (k+2)! - (k+2)+k!
Comment:: That does not seem to be working out.
Please check it out.
==============
Cheers,
Stan H.
===============================
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
a) Base cases n = 1 and n = 2 holds (you will see why I checked two cases). Assume that for some 
, that 
and 
. We look at the n+1th term:
(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 
. Then, by our hypothesis,
 + (n+1) \times (n+1)!)
! - 1 + (n+1) \times (n+1)!)
 \times (n+1)! - 1)
! - 1)
Therefore the statement holds for n+1, so by induction, we are done.
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