Difference between revisions of "TADM2E 2.33"

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On careful observation , one can see that the sum of any row is just  <math>3^<sup>(n-1)</sup></math>
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On careful observation , one can see that the sum of any row is just  <math>3^{n-1}</math>
 
this is the sum for the series . This can even be computed using a series as shown below  
 
this is the sum for the series . This can even be computed using a series as shown below  
  
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now a(mid) of the next line becomes the a(0) +a(1) +a(2)  that it this mid element is always the sum of the middle three elements
 
now a(mid) of the next line becomes the a(0) +a(1) +a(2)  that it this mid element is always the sum of the middle three elements
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'''Alternate explanation :'''
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Every element in the current row will be used 3 times in next row; once directly below it, and also in left and right element below it.
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So sum of elements in <math>(n+1)</math>th row will be 3*(sum of elements in <math>n</math>th row).
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Hence, S(n) = <math>3^{n-1}</math> (Base case : For 1st row, sum is 1)

Latest revision as of 12:07, 2 August 2020

On careful observation , one can see that the sum of any row is just $ 3^{n-1} $ this is the sum for the series . This can even be computed using a series as shown below

1=a0
1 1 1 
a1 a0 a2

now a(mid) of the next line becomes the a(0) +a(1) +a(2) that it this mid element is always the sum of the middle three elements


Alternate explanation :

Every element in the current row will be used 3 times in next row; once directly below it, and also in left and right element below it.

So sum of elements in $ (n+1) $th row will be 3*(sum of elements in $ n $th row).

Hence, S(n) = $ 3^{n-1} $ (Base case : For 1st row, sum is 1)