Variance of sum of $10$ random variables. Find the variance of sum of $10$ random variables if each has variance $5$ and if each pair has a correlation coefficient $.5$
Let $Y=X_1+X_2+X_3+\ldots+X_{10}$
I tried this problem by calculating variance of first $10$  random variables.
$V(Y)=50$. Then there will be $45$ pairs of covariance terms. 
Correlation coefficient $\rho=.5$ 
$Cov(X_i,X_j)=2.5 $
Which  gives variance $V(Y)=50+2.5=52.5$
Did i do everything right, please someone tell me.
 A: You did not make use of the number $45$, the number of pairs of covariance terms, in your final answer though you should.
Note that we have
\begin{align}\operatorname{Var}\left(\sum_{i=1}^nX_i\right) &= \sum_{i=1}^n\operatorname{Var}(X_i)+\color{blue}{2\sum_{i<j }} \operatorname{Cov(X_i,X_j)} \\
&=50+90 \times 2.5 \\
&=275\end{align}
A: Covariance is bilinear (and symmetric) so that:$$\mathsf{Var}(Y)=\mathsf{Covar}(Y,Y)=\sum_{i=1}^{10}\sum_{j=1}^{10}\mathsf{Covar}(X_i,X_j)$$
In your situation the variances are equal and the covariances are equal so we can go on by stating that this equals:$$=10\mathsf{Covar}(X_1,X_1)+90\mathsf{Covar}(X_1,X_2)=10\mathsf{Var}(X_1)+90\mathsf{Covar}(X_1,X_2)$$
Further $\mathsf{Var}(X_1)=5$ and $\mathsf{Covar}\left(X_{1},X_{2}\right)=\sqrt{\mathsf{Var}X_{1}}\sqrt{\mathsf{Var}X_{2}}\rho\left(X_{1},X_{2}\right)=5\times0.5=2.5$ 
So we end up with: $$\mathsf{Var}(Y)=10\times5+90\times2.5=275$$
A: $$Y=\sum_{n=1}^{10}X_n\to var(Y)=E(Y^2)-E^2(Y)\\=E(\sum_{n=1}^{10}X^2_n+\sum_{m,n=1\\m\ne n}^{10}X_nX_m)-(\sum_{n=1}^{10}E(X_n))^2\\=50+90\times 2.5-(\sum_{n=1}^{10}E(X_n))^2$$if we take $E(X_n)=0$ we will have:$$var(Y)=275$$
