# Difference between revisions of "TADM2E 2.33"

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Ccamenares (Talk | contribs) (For row 1, the sum is 1. So, the formula 3^n is incorrect. Instead, 3^(n-1) is correct.) |
m (Matt moved page User:TADM2E 2.33 to TADM2E 2.33 over redirect: revert) |
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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 | ||

Line 7: | Line 7: | ||

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 | ||

+ | |||

+ | |||

+ | '''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. | ||

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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)