Let $B_n(x)$ be the Bernoulli Polynomial.
1) Show for $n\neq 1$ is $B_n(1)=B_n(0) (=B_n)$.
2) Determine $B_1=B_1(0)$ and $B_1(1)$.
I've already tried just to plug in the values in different kinds of Bernoulli Polynomial representations (with exception of integral representation, because we didn't have this definition at lecture) but i don't get the right solution. I would be very grateful for any help.
Hint. Use the identity (see the generating function of Bernoulli_polynomials): $$\sum_{n=0}^{\infty}\frac{B_n(x)}{n!}u^n = \frac{u e^{xu}}{e^u - 1}.$$ Then, by replacing $x$ with $1-x$ and $u$ with $-u$, we get $$\sum_{n=0}^{\infty}\frac{B_n(1-x)}{n!}(-u)^n = \frac{-u e^{(1-x)(-u)}}{e^{-u} - 1} =\frac{u e^{xu}}{e^{u}-1}=\sum_{n=0}^{\infty}\frac{B_n(x)}{n!}u^n.$$ which implies the symmetry $(-1)^nB_n(1-x)=B_n(x)$.
Moreover take a look at Bernoulli odd numbers are 0 $B_{2n+1}=0,\;n>0$
