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^ | + | 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)
