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Patterns Portfolio

  • Date Submitted: 04/01/2010 07:03 PM
  • Flesch-Kincaid Score: 81.7 
  • Words: 814
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Truong Nguyen
Patterns of Recognition
This portfolio problem ask us to find the number of blocks in certain train patterns. It wants to know how many blocks are in the fourth train? How many blocks are in the eighth train? How can you tell how many blocks will be in a later train in the pattern?
For the diagram number one I got 1 block, for number two I got 5 blocks and for the diagram number three I got 10 blocks. If it went on to the fourth train, there will be 16 boxes. And there will be 36 blocks in the eighth train.
To find how many blocks are in the first pattern train, I counted it.   There was only 1 block in the first pattern.   I counted the second pattern and there were 5 more blocks; therefor the second pattern has 6 blocks total.   I counted the blocks again in the third pattern to find how many blocks are in it.   The third pattern has 5 more blocks added on to the second pattern.   Therefor, the third pattern has a total of 11 blocks.
I found the equation by observing the first three patterns given.   I know that the first part has only 1 block.   After the first pattern, there were 5 blocks added on to make the 2nd pattern.   After the second pattern, there were 5 more blocks added to make the third pattern.   From that observation, I know that there will be 5 more blocks added on each time to make the next pattern.   Because the first pattern is different from the other pattern in that it only has 1 block, I wrote 1 in the equation to represent the first pattern.   I represented the pattern number with the variable 'n'.   I know that each time the pattern number increases, you add 5 more blocks.   But since pattern number one, like I mentioned before, is different I subtract it from the 'n'.   Next I multiply the 'n-1' by five (the number of blocks added on more each time).   However, just because the pattern one is different doesn't mean we disregard it because pattern one has one block.   So I added 1 to [5x(n-1)].  
Now my equation is 1 + 5(n-1)....


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