Prove
             
P( n ):  1 + 2 + 3 + ... + n  =  n( n + 1 ) / 2

 

Here's another one:

Prove

P( n ):  1^2 + 2^2 + 3^2 + ... + n^2  =  n( n + 1 )( 2n + 1 ) / 6
 

1 )

ShowP( 1 )is true:

Stick a 1 in for all the n's and show it works.
 

P( 1 ):  1^2  =  1( 1 + 1 )( 2( 1 ) + 1 ) / 6  =  1 * 2 * 3 / 6  =  6 / 6  = 1
 

So,P( 1 )is true.

 

2 )

AssumeP( k )is true:

Stick a k in for all the n's and say it's true.

P( k ):  1^2 + 2^2 + 3^2 + ... + k^2  =  k( k + 1 )( 2k + 1 ) / 6

is true
 

 

3 )

ShowP( k )  -->  P( k + 1 )

Use P( k ) to show that P( k + 1 ) is true.

Write out your goal by sticking ( k + 1 ) in for all the n's...  Leave the k on the left side.

GOAL:  P( k + 1 ):  1^2 + 2^2 + 3^2 + ... + k^2 + ( k + 1 )^2  =  ( k + 1 )[ ( k + 1 ) + 1 ][ 2( k + 1 ) + 1 ] / 6

Start with the left side of the P( k + 1 ) and slowly turn it into the right side...  Don't confuse the monkey!
 

P( k + 1 ):  1^2 + 2^2 + 3^2 + ...+ k^2 + ( k + 1)^2  ...  Use P( k )!  =  k( k +1 )( 2k + 1 ) / 6  +  ( k + 1 )^2  ...  need that 6 and one big fraction  ...  k( k + 1 )( 2k +1 ) / 6  +  6( k + 1 )^2 / 6  =  k( k + 1 )( 2k + 1 ) + 6( k + 1 )^2 / 6


Look at the end goal...  What's in front?  ...  Factor it out!
 

=  ( k + 1 )[ k( 2k + 1 ) + 6( k + 1 ) ] / 6  ...  Simplify this end part...  =  ( k + 1 )[ 2k^2 + k + 6k + 6 ] / 6  ...  ( k + 1 )( 2k^2 + 7k + 6 ) / 6 ...  factor this...  it MUST work!  ...  You can look at the goal to cheat!  ...  = ( k + 1 )( k + 2 )( 2k + 3 ) / 6  ...  now, make these last parts look like the goal  ...  = ( k + 1 )[ ( k +1 ) + 1 ][ 2k + 2 + 1 ] / 6  = ( k + 1 )[ ( k + 1 ) + 1 ][ 2( k + 1 ) + 1 ] / 6

 

So,P( k )  -->  P( k + 1 )

 

4 )

Thus,P( n )is true.  []

 

OK, you need the practice...  Write this guy out without my [ boxed ]
comments.