(1)

\begin{align*} B_{n}\left(1-x\right) & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}\left(1-x\right)^{n-k}\\ & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}\sum_{j=0}^{n-k}C\left(n-k,j\right)\left(-x\right)^{j}\\ & =\sum_{k=0}^{n}\sum_{j=k}^{n}C\left(n,k\right)B_{k}C\left(n-k,j-k\right)\left(-x\right)^{j-k}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}C\left(n,k\right)B_{k}C\left(n-k,j-k\right)\left(-x\right)^{j-k}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}\frac{n!}{k!\left(n-k\right)!}\cdot\frac{\left(n-k\right)!}{\left(j-k\right)!\left(n-j\right)!}B_{k}\cdot\left(-x\right)^{j-k}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}\frac{n!}{j!\left(n-j\right)!}\cdot\frac{j!}{k!\left(j-k\right)!}B_{k}\cdot\left(-x\right)^{j-k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)\sum_{k=0}^{j}C\left(j,k\right)B_{k}\cdot\left(-x\right)^{j-k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)B_{j}\left(-x\right)\\ & =\left(-1\right)^{n}B_{n}\left(x\right) \end{align*}

(2)

ベルヌーイ多項式の母関数より、
\begin{align*} \sum_{k=0}^{\infty}\left\{ B_{k+1}\left(1+x\right)-B_{k+1}\left(x\right)\right\} \frac{t^{k+1}}{\left(k+1\right)!} & =\sum_{k=1}^{\infty}\left\{ B_{k}\left(1+x\right)-B_{k}\left(x\right)\right\} \frac{t^{k}}{k!}\\ & =\sum_{k=0}^{\infty}\left\{ B_{k}\left(1+x\right)-B_{k}\left(x\right)\right\} \frac{t^{k}}{k!}\\ & =\frac{te^{\left(1+x\right)t}}{e^{t}-1}-\frac{te^{xt}}{e^{t}-1}\\ & =\frac{te^{xt}}{e^{t}-1}\left(e^{t}-1\right)\\ & =te^{xt}\\ & =t\sum_{k=0}^{\infty}\frac{\left(xt\right)^{k}}{k!}\\ & =\sum_{k=0}^{\infty}\left(k+1\right)x^{k}\frac{t^{k+1}}{\left(k+1\right)!} \end{align*} 両辺の係数を比較すると、
\[ B_{k+1}\left(1+x\right)-B_{k+1}\left(x\right)=\left(k+1\right)x^{k} \] となる。
また\(k=-1\)に対しても成り立つ。
従って与式は成り立つ。

(2)-2

\begin{align*} B_{n+1}\left(1+x\right)-B_{n+1}\left(x\right) & =\left(-1\right)^{n+1}B_{n+1}\left(-x\right)-B_{n+1}\left(x\right)\\ & =\left(-1\right)^{n+1}\sum_{k=0}^{n+1}C\left(n+1,k\right)B_{k}\cdot\left(-x\right)^{n+1-k}-\sum_{k=0}^{n+1}C\left(n+1,k\right)B_{k}x^{n+1-k}\\ & =\sum_{k=0}^{n+1}\left(-1\right)^{k}C\left(n+1,k\right)B_{k}x^{n+1-k}-\sum_{k=0}^{n+1}C\left(n+1,k\right)B_{k}x^{n+1-k}\\ & =\sum_{k=0}^{n+1}\left(\left(-1\right)^{k}-1\right)C\left(n+1,k\right)B_{k}x^{n+1-k}\\ & =\sum_{k=0}^{\infty}\left(\left(-1\right)^{2k+1}-1\right)C\left(n+1,2k+1\right)B_{2k+1}x^{n+1-\left(2k+1\right)}\\ & =-2\sum_{k=0}^{\infty}C\left(n+1,2k+1\right)B_{2k+1}x^{n+1-\left(2k+1\right)}\\ & =-2C\left(n+1,1\right)B_{1}x^{n}\\ & =-2\left(n+1\right)\left(-\frac{1}{2}\right)x^{n}\\ & =\left(n+1\right)x^{n} \end{align*}

(3)

\begin{align*} \left(-1\right)^{n}B_{n}\left(-x\right) & =B_{n}\left(1+x\right)\\ & =B_{n}\left(x\right)+nx^{n-1} \end{align*}

(4)

略

(5)

(2)より、
\[ k^{n}=\frac{1}{n+1}\left(B_{n+1}\left(1+k\right)-B_{n+1}\left(k\right)\right) \] なので、
\begin{align*} \sum_{k=1}^{m}k^{n} & =\frac{1}{n+1}\sum_{k=1}^{m}\left(B_{n+1}\left(1+k\right)-B_{n+1}\left(k\right)\right)\\ & =\frac{1}{n+1}\left(B_{n+1}\left(1+m\right)-B_{n+1}\left(1\right)\right) \end{align*} となり、与式は成り立つ。