In the following triangle I need to find the sum of nth row. Is there a general formula for this? If yes, then please tell me.
- Triangle:
- Row 1: 1
- Row 2: 1 2 1
- Row 3: 1 3 6 3 1
- Row 4: 1 4 10 20 10 4 1
- Row 5: 1 5 15 35 70 35 15 5 1
- etc.
S(n) = sum of nth row.
Like, S(1) = 1 S(2) = 4 S(3) = 14 S(4) = 50 etc..
Find S(n)?
An attempt to show how the triangle is constructed
$$\binom 00$$ $$\binom 10\binom 21\binom 10$$ $$\binom20\binom31\binom 42\binom 31\binom 20$$ $$\binom 30\binom 41\binom 52\binom 63\binom 52\binom 41\binom 30$$
etc with the middle of row $r$ being $\binom {2r-2}{r-1}$ which is the V-shaped extracts fromPascal's Triangle noted in my comment below.
Another equivalent way, which may be more useful is $$\binom 00$$$$ \binom 11\binom 21\binom 11$$$$\binom 22\binom 32\binom 42\binom 32\binom 22$$
etc
If we count the rows starting at zero, the sum of row $r$ is $$2\left(\sum_{i=0}^{r-1} {r+i \choose i}\right) + {2r \choose r}$$ It starts off $1,4,14,50,182,672$, which picks out OEIS A051924 The last comment says these are the hook sums of Pascal's triangle and there is an explicit formula ${2n \choose n}-{2n-2 \choose n-1}$. This starts counting at $1$, not $0$