Difference between revisions of "2.37"
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(Created page with "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 b...") |
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| − | On careful observation , one can see that the sum of any row is just <math>3^{n-1}</math> | + | 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 |
1=a0 | 1=a0 | ||
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'''Alternate explanation :''' | '''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. | + | Every element in the current row will be used 3 times in the next row; once directly below it, and also in left and right element below it. |
So sum of elements in <math>(n+1)</math>th row will be 3*(sum of elements in <math>n</math>th row). | So sum of elements in <math>(n+1)</math>th row will be 3*(sum of elements in <math>n</math>th row). | ||
Hence, S(n) = <math>3^{n-1}</math> (Base case : For 1st row, sum is 1) | Hence, S(n) = <math>3^{n-1}</math> (Base case : For 1st row, sum is 1) | ||
| + | |||
Back to [[Chapter 2]] | Back to [[Chapter 2]] | ||
Latest revision as of 19:55, 10 September 2020
On careful observation, one can see that the sum of any row is just [math]\displaystyle{ 3^{n-1} }[/math] 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 the next row; once directly below it, and also in left and right element below it.
So sum of elements in [math]\displaystyle{ (n+1) }[/math]th row will be 3*(sum of elements in [math]\displaystyle{ n }[/math]th row).
Hence, S(n) = [math]\displaystyle{ 3^{n-1} }[/math] (Base case : For 1st row, sum is 1)
Back to Chapter 2
