what is wrong in here? a biased, consistent estimator of the mean given $N$ i.i.d samples is the sample mean. The variance of estimated mean is ${\mathop{\rm var}} (\hat \mu ) = \frac{{{\sigma ^2}}}{N}$ and the estimator is
$$\hat \mu  = \frac{1}{N}\sum\limits_{i = 1}^N {{x_i}} $$
However if I use the other formula for variance I get
${\mathop{\rm var}} \left( {\hat \mu } \right) = E{\left( {\hat \mu } \right)^2} - {\left( {E\left( {\hat \mu } \right)} \right)^2} = E{\left( {\frac{1}{N}\sum\limits_{i = 1}^N {{x_i}} } \right)^2} - {\left( {E\left( {\frac{1}{N}\sum\limits_{i = 1}^N {{x_i}} } \right)} \right)^2} = \frac{{{\sigma ^2}}}{N} - {\mu ^2}$
Why am I getting the extra additive term ,what did I do wrong?
 A: You have computed the expectation incorrectly.  The easiest way to obtain the result it to use the fact that the variance of a sum is the sum of variances when the summands are independent, i.e. if $x_i$ are iid with mean $\mu$ and variance $\sigma^2$, then
$$\text{Var}\left(\frac{1}{N}\sum_{i=1}^N x_i\right)=\frac{1}{N^2}\sum_{i=1}^N \text{Var}(x_i)=\frac{1}{N^2}\sum_{i=1}^N \sigma^2=\frac{1}{N^2}N\sigma^2=\frac{\sigma^2}{N}.$$
Using your approach correctly would you get you there as well, albeit with a bit more work:
$$E\left[\left(\frac{1}{N}\sum_{i=1}^N x_i\right)^2\right]-\left(E\left[\frac{1}{N}\sum_{i=1}^N x_i\right]\right)^2\\
=\frac{1}{N^2}E\left[\sum_{i=1}^Nx_i^2+2\sum_{i=1}^N\sum_{j=i+1}^Nx_ix_j\right]-\mu^2\\
=\frac{1}{N^2}\left[\sum_{i=1}^NE\left[x_i^2\right]+2\sum_{i=1}^N\sum_{j=i+1}^N E\left[x_ix_j\right]\right]-\mu^2\\
=\frac{1}{N^2}\left[\sum_{i=1}^NE\left[x_i^2\right]+2\sum_{i=1}^N\sum_{j=i+1}^N E\left[x_i\right]E\left[x_j\right]\right]-\mu^2\quad \left(x_i. \text{ indep.}\right)\\
=\frac{1}{N^2}\left[\sum_{i=1}^N \left\{\mu^2+\sigma^2\right\}+2\sum_{i=1}^N\sum_{j=i+1}^N \mu^2\right]-\mu^2\quad \left(x_i \text{ identically distrib.}\right)\\
=\frac{1}{N^2}\left[N(\mu^2+\sigma^2)+(N^2-N)\mu^2\right]-\mu^2\\
=\frac{\sigma^2}{N}
$$
A: Like I said in a comment, your calculation of the first sum is wrong. Using the formula
$$(\sum_{i=1}^n x_i)^2=\sum_{i=1}^n x_i^2 + 2\sum_{j=1}^n\sum_{i=1}^{j-1}x_ix_j$$
You get
$$\mathbb{E}[(\sum_{i=1}^n x_i)^2]=\mathbb{E}[\sum_{i=1}^n x_i^2 + 2\sum_{j=1}^n\sum_{i=1}^{j-1}x_ix_j]$$
$$=n\mathbb{E}[x_1^2]+2\sum_{j=1}^n\sum_{i=1}^{j-1}\mathbb{E}[x_ix_j]=n\mathbb{E}[x_1^2]+2\sum_{j=1}^n\sum_{i=1}^{j-1}\mu^2$$
$$=n\mathbb{E}[x_1^2]+2\mu^2\sum_{j=1}^n(j-1)=(n\mathbb{E} -n\mu^2) 
 +n\mu ^2 +2\mu^2\sum_{j=1}^n(j-1)$$
$$=n\sigma^2+n\mu^2+2\mu^2\frac{n(n-1)}{2}=n\sigma^2+n^2\mu^2$$
so
$$\mathbb{E}[(1/n\sum_{i=1}^n x_i)^2]=\frac{\sigma^2}{n}+\mu^2$$
So the formula checks out
