Find the value of $n$. Write the value of $n$ if the sum of n terms of the series $1+3+5+7...n =n^2$.
I'm not getting the right value if I proceed with the general formula for finding sum of n terms of a arithmetic series. The general summation formula for arithmetic series is $\frac{n(2a+(n-1)d)}{2}$, where $a$ is the first term, $d$ is the common difference and $n$ is the number of terms.

If we add $2+4+...+(n-1) = 2(1+2+3+...+\frac{n-1}{2})$ to both sides of the equation, we get 
$$1+2+3+...+n = n^2+2\bigg(1+2+3+...+\frac{n-1}{2}\bigg)$$
which can be written as:
$$\frac{n(n+1)}{2} = n^2+2\bigg[\frac{\big(\frac{n-1}{2}\big)\big(\frac{n+1}{2}\big)}{2} \bigg]$$
that is
$$\frac{n^2+n}{2} = n^2+\frac{n^2-1}{4} \implies n^2+n = \frac{5n^2-1}{2} \implies 3n^2-2n-1 = 0 \implies n = 1$$
because $n$ is not a negative number.
