Binomial Summations 2

I have a method which uses only the identity that $\displaystyle \sum_{k=0}^n \dbinom{n}{k} = 2^n$ . We can rewrite the summation in the following array:

$\begin{array} {c c c c} \dbinom{13}{0}+\\ \dbinom{13}{0}+ & \dbinom{13}{1}+\\ \dbinom{13}{0}+ & \dbinom{13}{1}+ & \dbinom{13}{2}+\\ \vdots & & & \ddots \end{array}$

Let the first row be Row 1, the second row be Row 2, etc. Notice that the sum of Row $n$ and Row $14-n$ have the same sum. For example, Row 1 is $\dbinom{13}{0}$ and Row 13 is $\dbinom{13}{0}+\dbinom{13}{1}+\ldots+\dbinom{13}{12}$ , converting $\dbinom{13}{0}$ to $\dbinom{13}{13}$ for Row 1, we get that the sum will have the exact terms as Row 14. We can do this similarly for Rows 2, 3, 4, 5 and 6. If we add Row 7 to itself, we can also do this, so the sum of Row 7 is half that of Row 14. The sum of Row 14 is $2^{13}$ , so the total sum (remembering to add Row 14 as well) is $\dfrac{15}{2} \times 2^{13} = 61440$ .

We can generalise this sum to $(n+2) 2^{n-1}$ , where $n=13$ in this case.

$\begin{aligned} S & = \sum_{j=0}^{13} \sum_{i=0}^j {13 \choose i} \\ & = \sum_{j=0}^{13} {13 \choose j}(14-j) \\ & = 14 \sum_{j=0}^{13} {13 \choose j} - \sum_{j=0}^{13} {13 \choose j}j \\ & = 14\cdot 2^{13} - \sum_{j=0}^{13} \frac {13!\cdot j}{j!(13-j)!} \\ & = 14\cdot 2^{13} - \sum_{j=0}^{12} \frac {13\cdot 12!}{j!(12-j)!} \\ & = 14\cdot 2^{13} - 13 \sum_{j=0}^{12} {12 \choose j} \\ & = 14\cdot 2^{13} - 13 \cdot 2^{12} \\ & = \boxed{61440} \end{aligned}$