Posted by Francesca on Monday, March 21, 2011 at 12:42pm.
Hello?
The next step is to assume the proposition is true for n.
The task is to show that if the proposition is true for n, then it would be true for n+1. Once that is established, then the proof is complete.
6|n^3+5n => 6|1+5 is true for n=1
Assume 6|N^3+5 is true for n, then
for n+1
(n+1)^3+5(n+1)
=n^3+5n +3n^2+3n+6
=n^3+5n +3n(n+1) + 6
We now examine the three terms:
n^3+5n is divisible by 6 by initial assumption.
6 is divisible by 6.
3n(n+1) falls into two cases:
1. n is odd, then n+1 is even, therefore 6 divides 3*(n+1)
2. n is even, then 6 divides 3n.
Since all three terms are divisible by 6, we only have to extract the factor of 6 from each term and declare the expression (n+1)^3+5(n+1) is also divisible by 6.
By the principle of mathematical induction, the proposition is proved. QED.
Thank you so much for your response! But I have completed that particular question. However, can you please help with this one? I am confused. . .
Use mathematical induction to establish the following formula.
n
Σ i² / [(2i-1)(2i+1)] = n(n+1) / 2(2n+1)
i=1
Thanks for any helpful replies :)
Any suggestions?
There are three steps:
1. Basis:
Test case for n=1 (or any other finite number):
n
Σ i² / [(2i-1)(2i+1)] = n(n+1) / 2(2n+1)
i=1
for n=1,
Left hand side=1/[(2*1-1)(2*1+2)=1/3
Right hand side=1(1+1)/[2(2*1+1)]=1/3
So formula is established for n=1.
2. Assume
formula is valid for case n.
3. Show that formula is valid for case n+1.
Left hand side:
n+1
Σ i² / [(2i-1)(2i+1)]
i=1
=
n
Σ i² / [(2i-1)(2i+1)]
i=1
+ (n+1)²/[(2*(n+1)-1)(2*(n+1)+2)]
=
n(n+1) / 2(2n+1) + (n+1)²/[(2*(n+1)-1)(2*(n+1)+2)]
=
[(n^2+n)*(2n+3)+2(n+1)^2] / [2(2n+1)(2n+3)]
=(n+1)(n+2)(2n+1) / [2(2n+1)(2n+3)]
=(n+1)((n+1)+1))/[2(2(n+1)+1)]
=m(m+1)/[2(2m+1)]
Which is precise the right hand side with m replacing n+1.
QED