# 2.37

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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)

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