This is the question, I have solved it but I need someone to double check my solution. Question: Find the temperature $u(x, t)$ in a rod of length $L$ if the initial temperature is $f(x)$ throughout and if the ends $x=0$ and $x=L$ are insulated. $F(x)= x, 0$
Solution:
For an insulated rod the solution $$ X(x,t)= \frac{a_0}{2}+\sum B_0 \frac{\cos(n\pi)x}{Le}−\frac{n^2\pi^2\alpha^2}{L}t $$ I found $a_0= 1$ and $$ B_n= −\frac{2}{n\pi}\sin\left(\frac{n\pi}{2}\right)+\frac{2}{n\pi}2\cos\left(\frac{n\pi}{2}\right)−\left(\frac{2}{n\pi}\right)^2 $$ then just plug in the coefficients into the sum. I am just not sure if these are the correct values for the $a_0$ and $B_0$ coefficients.
(As another user pointed out), your formula for the solution should be written:
$$ X(x,t)= \frac{a_0}{2}+\sum B_{\mathbf{n}} \frac{\cos(n\pi x)}{L}\mathbf{e}^{−\frac{n^2\pi^2\alpha^2}{L}t} $$
Also, you don't have the correct formulas for the coefficients.
Recall $a_0 = (2/L)\int_0^L f(x) dx, \ $ $B_n = (2/L)\int_0^L \cos(n\pi x / L) f(x) dx$. So, with $f(x) = x$, we have
$$a_0 = (2/L)\int_0^L x dx = L$$
$$b_n = (2/L)\int_0^L \cos(n\pi x / L) x dx = (2/L)[x (L/n\pi) \sin(n \pi x / L) + (L/n\pi )^2 \cos(n \pi x / L)]_{x=0}^L$$ $$ = 2L/(n\pi)^2(\cos(n\pi) - 1) $$ ( = 0 if n even, $-4L/(n\pi)^2$ if n odd)