$X_t=\int_{0}^{t}(a_{0}+a_{1}\frac{u}{t}+\ldots+a_{n}\frac{u^{n}}{t^{n}})dB(u)$ is a Brownian motion for suitable non-zero constants $a_0,\ldots,a_n$ Let $B(t)$ be brownian motion. Show that for any integer $n \geq 1$, there
exist nonzero constants $a_{0},\ldots,a_{n}$ such that $X_{t}=\int_{0}^{t}(a_{0}+a_{1}\frac{u}{t}+\ldots+a_{n}\frac{u^{n}}{t^{n}})dB(u)$ is Brownian motion
 A: There is the following characterization of (one-dimensional) Brownian motion:

Let $(X_t)_{t \geq 0}$ be a stochastic process such that the random vector $(X_{t_1},\ldots,X_{t_n})$ is a Gaussian random variable with mean $0$ and covariance matrix $(t_j \wedge t_k)_{j,k=1,\ldots,n}$ for any $t_1,\ldots,t_n>0$, $n \in \mathbb{N}$. If $(X_t)_{t \geq 0}$ has continuous sample paths, then $(X_t)_{t \geq 0}$ is a Brownian motion.

For a proof see e.g. Brownian motion - An Introduction to Stochastic Processes by René Schilling & Lothar Partzsch, chapter 2. 

Obviously,
$$X_t = \int_0^t \sum_{k=0}^n a_k \frac{u^k}{t^k} \, dB_u$$
has continuous sample paths and, since the integrand is deterministic, the process is a Gaussian process. Now fix $t_1,\ldots,t_n>0$ for some $n \in \mathbb{N}$. Because of the martingale property of stochastic integrals (with respect to Brownian motion), we find that the random vector $Y := (X_{t_1},\ldots,X_{t_n})$ has mean $0$. Moreover, we already know that $Y$ is Gaussian. Consequently, it suffices to calculate the covariance matrix. For any $s<t$, we have by Itô's isometry
$$\mathbb{E}(X_t X_s) = \int_0^s \left( \sum_{k=0}^n a_k \frac{u^k}{t^k} \right) \left( \sum_{j=0}^n a_j \frac{u^j}{s^j} \right) \, du.$$
Hence,
$$\mathbb{E}(X_t X_s) = \sum_{k,j=0}^n a_k a_j \frac{1}{t^k} s^{k+1} \frac{1}{j+k+1} = s \sum_{k=0}^n c_k a_k \left( \frac{s}{t} \right)^k$$
for $c_k:= \sum_{j=0}^n a_j \frac{1}{j+k+1}$. Consequently, we find that
$$\mathbb{E}(X_t X_s) = s = \min\{s,t\}$$
if $c_k=0$ for all $k \in \{1,\ldots,n\}$ and $c_0 a_0 = 1$. That we can choose $a_0,\ldots,a_n$ in such a way that these conditions are satisfied follows like this: Set $c_0 := 1$ and $a_0 := 1$, then, clearly, the condition $c_0 a_0 = 1$ is satisfied. Now, for $k=1$, we have
$$c_1 = \frac{a_0}{2} + \frac{a_1}{3} = \frac{1}{2} + \frac{a_1}{3} \stackrel{!}{=}  1,$$
i.e. $$a_1 = 3 \left(1- \frac{1}{2} \right) = \frac{3}{2}.$$
Proceeding iteratively, we find that we can choose $a_0,\ldots,a_n >0$ in such a way that the conditions on $c_0,\ldots,c_n$ are satisfied.
This gives us that the covariance matrix of the random vector $Y$ equals (for the $a_0,\ldots,a_n$ we have just chosen) the matrix $$C = (t_j \wedge t_k)_{j,k=1,\ldots,n}$$
Applying the result from the beginning of my answer proves that the so-defined process is a Brownian motion.
