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):
|
|
|
