Find the greatest integer less than or equal to $(\sqrt 2 + 1)^6$ The greatest integer less than or equal to $(\sqrt 2  + 1)^7$ is _____.
I have solved the same for finding the greatest integer less than or equal to $(\sqrt 2  + 1)^6$ and my detailed solution is elaborated below

Can we find the same when power is 7 if so then how do we do that.
 A: One approach to avoid computing all those binomial coefficients is to use that: $$x_n=(1+\sqrt{2})^n+(1-\sqrt2)^n$$ satisfies $$x_{n+2}=2x_{n+1}+x_n$$ With $x_0=x_1=2.$
And, since $-1<1-\sqrt2<0,$ we get,  for any $n>0,$ $$\left\lfloor (1+\sqrt 2)^n\right\rfloor=\begin{cases}x_n&n\text{ odd}\\x_n-1&n\text{ even}\end{cases}$$
This gives $$x_2=6\\x_3=14\\x_4=34\\x_5=82\\x_6=198\\x_7=478\\\vdots$$

Amusingly, since $$\frac{1}{2}1^n+\frac{1}{2}(-1)^n=\begin{cases}1&n\text{ even}\\0&n\text{ odd}\end{cases}$$
You can actually get a recurrence for $y_n=\left\lfloor(1+\sqrt 2)^n\right\rfloor$ as: $$y_{n+4}=2y_{n+3}+2y_{n+2}-2y_{n+1}-y_n$$
A: $(\sqrt2+1)^7+(1-\sqrt2)^7=2(1^7+\binom72\sqrt2^21^5+\binom74\sqrt2^41^3+\binom76\sqrt2^61^1)=478$, and $-1<(1-\sqrt2)^7<0$.
So the answer is 478.
A: I will try to add on with same end result but in a simpler manner.
$$
(\sqrt{2}+1)^6=\mathrm{A}
$$
[A] is the greatest integer less than $\mathrm{A}$.
Let $f$ be some unknown fraction.
Hence, we can write $[\mathrm{A}]+\mathrm{f}=\mathrm{A}$.
Now, $(\sqrt{2}-1)<1$
$$
(\sqrt{2}-1)^6<1
$$
Let, $(\sqrt{2}-1)^6$ be denoted by $f_1$.
So, $\quad(\sqrt{2}+1)^6+(\sqrt{2}-1)^6=[\mathrm{A}]+\mathrm{f}+\mathrm{f}_1$
$$
\Rightarrow 2\left({ }^6 \mathrm{C}_0 \cdot(\sqrt{2})^6+{ }^6 \mathrm{C}_2 \cdot(\sqrt{2})^4+{ }^6 \mathrm{C}_6\right)=[\mathrm{A}]+\mathrm{f}+\mathrm{f}_1
$$
$$
\Rightarrow 2(8+60+31)=[\mathrm{A}]+\mathrm{f}+\mathrm{f}_1
$$
$$
\Rightarrow 198=[\mathrm{A}]+\mathrm{f}+\mathrm{f}_1
$$
Now, $0<\mathrm{f}<1$
$0<\mathrm{f}_1<1$
$0<\mathrm{f}+\mathrm{f}_1<2$
But as $198-[\mathrm{A}]=\mathrm{f}+\mathrm{f}_1$, So, $f+\mathrm{f}_1$ has to be an integer.
As, the only integer in $(0,2)$ is 1 , hence $\mathrm{f}+\mathrm{f}_1=1$
[A] = 197
Just like the other answers posted earlier.
Thank you
