Given that $ U_n=[x(1-x)]^n$ and $n\geq2$,$V_n=\int_{0}^{1}e^xU_ndx$,prove that $V_n+2n(2n-1)V_{n-1}-n(n-1)V_{n-2}=0$ Given that $ U_n=[x(1-x)]^n$ and $n\geq2$,$V_n=\int_{0}^{1}e^xU_ndx$,prove that $V_n+2n(2n-1)V_{n-1}-n(n-1)V_{n-2}=0$
I tried to solve it by integration by parts,taking $U_n$ as first function and $e^x$ as second function.
$V_n=-\int_{0}^{1}n[x(1-x)]^{n-1}(1-2x)e^x dx=-\int_{0}^{1}nU_{n-1}(1-x-x)e^x dx$
I could not proceed further,how can i reach the desired result.Please help.
 A: You can integrate $U_n \cdot e^x$, $U_{n-1} \cdot e^x$ and $xU_{n-1} \cdot e^x$ by parts then look for a suitable linear combination. Alternatively, you can find directly that
$$
(U_n +2n(2n-1)U_{n-1}-n(n-1)U_{n-2})\cdot e^x = 
\Big((U_n-n(1-2x)U_{n-1})\cdot e^x\Big)'.
$$
A: The factors $n(n-1)$ and $2n(2n-1)$ indicate that we could try integration by parts twice.

We obtain for $n\geq 2$
  \begin{align*}
V_n&=\int_0^1e^xU_n(x)dx\\
&=\int_0^1e^xx^n(1-x)^ndx\\
&=\left.e^xx^n(1-x)^n\right|_0^1-\int_0^1e^x\left[nx^{n-1}(1-x)^n-nx^n(1-x)^{n-1}\right]dx\tag{1}\\
&=-n\int_0^1e^xx^{n-1}(1-x)^ndx+n\int_0^1e^xx^{n}(1-x)^{n-1}dx\\
&=n\int_0^1e^x\left[(n-1)x^{n-2}(1-x)^n-nx^{n-1}(1-x)^{n-1}\right]dx\tag{2}\\
&\qquad-n\int_0^1e^x\left[nx^{n-1}(1-x)^{n-1}-(n-1)x^n(1-x)^{n-2}\right]dx
\end{align*}

Comment:


*

*In (1) and (2) we do integration by parts


The next step is to collect the summands in (2). Since the integral is a linear operator we can focus on the polynomial parts. We also conveniently use $U_n(x)$ as shorthand for $x^n(1-x)^n$.

We obtain from (2)
\begin{align*}
n&(n-1)(1-x)^2U_{n-2}(x)-n^2U_{n-1}(x)-n^2U_{n-1}(x)+n(n-1)x^2U_{n-2}(x)\\
&=n(n-1)(1-2x+2x^2)U_{n-2}(x)-2n^2U_{n-1}(x)\\
&=n(n-1)(1-2x(1-x))U_{n-2}(x)-2n^2U_{n-1}(x)\\
&=n(n-1)U_{n-2}(x)-2n(n-1)U_{n-1}(x)-2n^2U_{n-1}(x)\\
&=n(n-1)U_{n-2}(x)-2n(2n-1)U_{n-1}(x)\tag{3}
\end{align*}

Now it's time to harvest.

We obtain with (2) and (3)
  \begin{align*}
V_n&=-2n(2n-1)\int_0^1e^xU_{n-1}(x)\,dx+n(n-1)\int_{0}^1e^xU_{n-2}(x)\,dx\\
&=-2n(2n-1)V_{n-1}+n(n-1)V_{n-2}
\end{align*}
and the claim follows.

