Prove

Here's another one:
Prove

| So, |
 |
is true. |
|
 |
Assume |
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is true: |


is true
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 |
Show |
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![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](images/09-sequences-and-series-66.gif)



![= ( 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](images/09-sequences-and-series-70.gif)
| So, |
 |
| OK,
you need the practice... Write this guy out
without my |
![[ boxed ]](images/09-sequences-and-series-75.gif) |
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comments. |
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