6. Given a positive integer $n(n \geqslant 5)$, construct an inverted triangular number table as follows (see Figure 2): In the first row, write the numbers $1,2, \cdots, n$ in sequence. Below each pair of adjacent numbers in the row above, write their sum to form the next row (which has one fewer number than the row above). Continue this process until the last row (the $n$-th row) contains only one number, denoted as $f(n)$. Then $f(n)=$
$$
6.2^{n-2}(n+1) \text {. }
$$

Without loss of generality, let the first row be $a_{1}, a_{2}, \cdots, a_{n}$ forming an arithmetic sequence with a common difference of $d$. Then the second row is $a_{1}+a_{2}, a_{2}+a_{3}, \cdots, a_{n-1}+a_{n}$, also forming an arithmetic sequence with a common difference of $2d$; the third row is $a_{1}+2a_{2}+a_{3}, a_{2}+2a_{3}+a_{4}, \cdots, a_{n-2}+2a_{n-1}+a_{n}$, also forming an arithmetic sequence with a common difference of $4d \cdots \cdots$ Therefore, each row in Figure 2 forms an arithmetic sequence, with common differences being $1, 2, 2^{2}, \cdots, 2^{n-1}$.

Let the $j$-th number in the $i$-th row of Figure 2 be $f(i, j)$, then the number we are looking for is $f(n, 1)$. Hence,
$$
\begin{array}{l}
f(n, 1)=f(n-1,1)+f(n-1,2) \\
=f(n-1,1)+\left(f(n-1,1)+2^{n-2}\right) \\
=2 f(n-1,1)+2^{n-2} \\
=2(f(n-2,1)+f(n-2,2))+2^{n-2} \\
=2\left(f(n-2,1)+f(n-2,1)+2^{n-3}\right)+2^{n-2} \\
=2^{2} f(n-2,1)+2 \times 2^{n-2} \\
=\cdots=2^{n-1} f(1,1)+2^{n-2}(n-1) \\
=2^{n-1}+2^{n-2}(n-1)=2^{n-2}(n+1) .
\end{array}
$$