One small thing I noticed here is that 以上 (and presumable 以下 as well) are inclusive comparisons, so in a math context 2 以上 is \geq2 not >2.

##
English version of my proof for part (1)

\begin{align*}(2^n-1)(2^n-1) &= \left(\sum_{k=0}^{n-1}2^k\right)^2\\
&=\sum_{0\le j,k\le n-1}2^j\cdot2^k\\
&=2\cdot\left(\sum_{0\le j<k\le n-1}2^j\cdot2^k\right)+\sum_{k=0}^{n-1}2^{2k}\\
&=2\cdot a_{n,2}+\sum_{k=0}^{n-1}2^{2k}\\
a_{n,2}&=\frac{(2^n-1)^2-\sum_{k=0}^{n-1}2^{2k}}{2}
\end{align*}

Let

b_n=\sum_{j=0}^n2^{2j}

(In line math doesn’t seem to work inside of a details thing)

Then

\begin{align*}
3\cdot b_n&=\sum_{j=0}^n3\cdot2^{2j}\\
&=\sum_{j=0}^n\left(2\cdot2^{2j}+2^{2j}\right)\\
&=\sum_{j=0}^n\left(2^{2j+1}+2^{2j}\right)\\
&=\sum_{j=0}^{2n}2^j\\
&=2^{2n+2}-1\\
b_n&=\frac{2^{2n+2}-1}{3}
\end{align*}

Substituting into our original equation,

\begin{align*}
a_{n,2}&=\frac{(2^n-1)^2-\sum_{k=0}^{n-1}2^{2k}}{2}\\
&=\frac{(2^n-1)^2-b_{n-1}}{2}\\
&=\frac{(2^n-1)^2-\frac{2^{2n}-1}{3}}{2}\\
&=\frac{3(2^n-1)^2-2^{2n}+1}{6}\\
&=\frac{3\cdot2^{2n}-3*2^{n+1}+1-2^{2n}+1}{6}\\
&=\frac{2\cdot2^{2n}-6\cdot2^n+2}{6}\\
a_{n,2}&=\boxed{\frac{2^{2n}+2}{3}-2^n}
\end{align*}

I wrote this with very few actual words, so hopefully making a Japanese version won’t be too bad, but my vocabulary and grammar knowledge in this area is severely lacking. There’s definitely a specific way of writing for math proofs in English, so I’m assuming Japanese is the same, with plenty of set phrases you wouldn’t see elsewhere. I am going to try to translate this based on some language used in the explanation, but I’m especially unsure about how to declare my definition for b. I’d welcome any suggestions on both Japanese tips and parts of the proof I should explain more. I tend to write very concise proofs, and it’s also been 6 or 7 years since I had to write a math proof for someone else to read.

For my own purposes, I also reworded the problem statement in English, involving literally no actual translation. (spoiled instead of collapsed so I can use inline math)

Let n,k be integers with 1\le k \le n. For a given n, choose k distinct

2^m\space\space(m=0,1,2,\ldots,n-1)

Then consider the sum of all _nC_k such selections, and call that a_{n,k}. For example,

a_{4,3}=2^0\cdot2^1\cdot2^2+2^0\cdot2^1\cdot2^3+2^0\cdot2^2\cdot2^3+2^1\cdot2^2\cdot2^3

(1) For n\ge2, what is a_{n,2}?

(2) For n\ge1, think about the following function of x

f_n(x)=1+a_{n,1}x+a_{n,2}x^2+\cdots+a_{n,n}x^nf_n(x)=1+a_{n,1}x+a_{n,2}x^2+\cdots+a_{n,n}x^n

What are \frac{f_{n+1}(x)}{f_n(x)} and \frac{f_{n+1}(x)}{f_n(2x)} in terms of x?

(3) Express \frac{a_{n+1,k+1}}{a_{n,k}} in terms of n,k.

I also wrote a proof of part (2) consisting of only equations, but I would probably need to actually prove the first statement I make there.

##
Very brief part 2 proof

\begin{align*}
f_n(x)&=\prod_{j=0}^{n-1}(1+2^jx)\\
f_{n+1}(x)&=\prod_{j=0}^{n}(1+2^jx)\\
\frac{f_{n+1}(x)}{f_n(x)}&=\boxed{1+2^nx}\\
f_{n}(2x)&=\prod_{j=0}^{n-1}(1+2^j\cdot2x)\\
&=\prod_{j=1}^{n}(1+2^jx)\\
\frac{f_{n+1}(x)}{f_n(2x)}&=\boxed{1+x}
\end{align*}