2.37

On careful observation, one can see that the sum of any row is just ${\displaystyle 3^{n-1}}$ 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 the next row; once directly below it, and also in left and right element below it.

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

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

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