6. Let the integer part and the fractional part of $(5 \sqrt{2}+7)^{2 n+1}(n \in \mathbf{N})$ be $I$ and $F$, respectively. Then the value of $F(I+F)$ is ( ).
(A) 1
(B) 2
(C) 4
(D) a number related to $n$
6. A.

Since $5 \sqrt{2}-7 \in(0,1)$ and
$(5 \sqrt{2}+7)^{2 n+1}(5 \sqrt{2}-7)^{2 n+1}=1$,
also $(5 \sqrt{2}+7)^{2 n+1}=I+F$.
Therefore, $(5 \sqrt{2}-7)^{2 n+1}=\frac{1}{I+F}$.
By the binomial theorem, $(5 \sqrt{2}+7)^{2 n+1}-(5 \sqrt{2}-7)$
is an integer. That is, $I+F-\frac{1}{I+F}$ is an integer, and since $I$ is an integer, hence $F$ and $\frac{1}{I+F}$ are integers.

Since $0<F<1$, so $F=\frac{1}{I+F}$, then $(1+F)=1$.