恒等式结论

引理： ∑ ϵ k ∈ { − 1 , 1 } , 1 ≤ k ≤ n ( ∑ k = 1 n ϵ k a k ) 2 = 2 n ∑ k = 1 n a k 2 \sum_{\epsilon_k\in \{-1,1\},1\le k\le n}(\sum^n_{k=1}\epsilon_ka_k)^2=2^n\sum^n_{k=1}a_k^2 ϵk​∈{−1,1},1≤k≤n∑​(k=1∑n​ϵk​ak​)2=2nk=1∑n​ak2​
证明：$$n=1时，LHS=2a_1^2，RHS=2a_1^2，成立$$
$$设n时成立，证明n+1时情况：$$
∑ ϵ k ∈ { − 1 , 1 } , 1 ≤ k ≤ n + 1 ( ∑ k = 1 n + 1 ϵ k a k ) 2 \sum_{\epsilon_k\in \{-1,1\},1\le k\le n+1}(\sum^{n+1}_{k=1}\epsilon_ka_k)^2 ϵk​∈{−1,1},1≤k≤n+1∑​(k=1∑n+1​ϵk​ak​)2
= ∑ ϵ k ∈ { − 1 , 1 } , 1 ≤ k ≤ n + 1 ( ( ∑ k = 1 n ϵ k a k ) 2 + 2 ( ∑ k = 1 n ϵ k a k ) ∗ ϵ n + 1 a n + 1 + a n + 1 2 ) =\sum_{\epsilon_k\in \{-1,1\},1\le k\le n+1}((\sum^n_{k=1}\epsilon_ka_k)^2+2(\sum^n_{k=1}\epsilon_ka_k)*\epsilon_{n+1}a_{n+1}+a_{n+1}^2) =ϵk​∈{−1,1},1≤k≤n+1∑​((k=1∑n​ϵk​ak​)2+2(k=1∑n​ϵk​ak​)∗ϵn+1​an+1​+an+12​)
= 2 ∑ ϵ k ∈ { − 1 , 1 } , 1 ≤ k ≤ n + 2 ∑ ϵ n + 1 ∈ { − 1 , 1 } ϵ n + 1 a n + 1 ∑ ϵ k ∈ { − 1 , 1 } , 1 ≤ k ≤ n ϵ k a k + 2 n + 1 × a n + 1 2 =2\sum_{\epsilon_k\in \{-1,1\},1\le k\le n}+2\sum_{\epsilon_{n+1}\in\{-1,1\}}\epsilon_{n+1}a_{n+1}\sum_{\epsilon_k\in \{-1,1\},1\le k\le n}\epsilon_ka_k+2^{n+1}\times a_{n+1}^2 =2ϵk​∈{−1,1},1≤k≤n∑​+2ϵn+1​∈{−1,1}∑​ϵn+1​an+1​ϵk​∈{−1,1},1≤k≤n∑​ϵk​ak​+2n+1×an+12​
$$=2^{n+1}\sum _{k=1}^n{a}_k^2+2^{n+1}\times a_{n+1}^2=RHS证毕$$