Let $(M^n,g)$ be an oriented Riemannian $n$- manifold and $g$ is a Riemannian metric on $M$ , $\mathrm{d}\sigma$ is Riemannian volume form on $S^{n-1}$ and $\text{Vol}(S^{n-1})$ is volume of $S^{n-1}$.
I have proved that the average of a symmetric 2-tensor $\alpha(V,V)$ on all unit vectors $V\in S^{n-1}\subset T_pM^n$ is $\dfrac{1}{n}$ of trace of $\alpha$. i.e.
$\dfrac{1}{n}\text{trace}_g(\alpha)=\dfrac{1}{\text{Vol}(S^{n-1})}\displaystyle\int_{S^{n-1}}\!\alpha(V,V)\mathrm{d}\sigma$.
From this, How can I show that for any unit vector $U$ that $\dfrac{1}{n-1}\mathrm{Ric}(U,U)$ (Ricci tensor) is the average of the sectional curvatures of planes containing the vector $U$.
Let $\mathrm{R}$ is Riemannian tensor and $\mathrm{K}$ is sectional curvature. I defined that $\alpha(Y,Z):=\mathrm{R}(U,Y,Z,U)$.It is clear that $\alpha$ is symmetric 2-tensor. Now $\alpha(V,V)=\mathrm{K}(U,V)$.How does it Work? What's your ideas?
First of all you need to make sense of "averaging along all planes containing $U$". This can be done: Let $U^\perp \subset T_pM$. Then all planes containing $U$ are in one-to-one correspondence with the unit sphere $S^{n-2} \subset U^\perp$.
Thus the average $A$ is given by
$$A = \frac{1}{Vol(S^{n-2})} \int_{S^{n-2}} \ \mathrm R(U, V, U, V) d\sigma = \frac{1}{Vol(S^{n-2})} \int_{S^{n-2}} \ \alpha(V, V) d\sigma. $$
Now from what you have proved (well, not exactly what you proved), we have
$$\begin{split} \frac{1}{n-1} Ric (U, U) &= \frac{1}{n-1} \left(R(U, U, U, U) + \sum_{i=2}^n R(U, e_i, U,e_i)\right) \\ &=\frac{1}{n-1} \sum_{i=2}^n R(U, e_i, U,e_i)\\ &= \frac{1}{n-1} tr (\alpha|_{U^\perp})\\ &= A \end{split}$$