I have to find whether the function $\mathrm f(x)$ defined as $$f(x)=\begin{cases}\cfrac{x+3x^2+5x^3+\cdots+(2n-1)x^n-n^2}{(x-1)}, \;\;&\text{for $x\ne1$}\\[2ex]\cfrac{n(n^2-1)}{3} &\text{for $x=1$}\end{cases}$$
is continuous at $x=1$.
Now, I need to evaluate $$\lim_{x\rightarrow1}\frac{x+3x^2+5x^3+\cdots+(2n-1)x^n-n^2}{(x-1)}$$
The series $S=x+3x^2+5x^3+\cdots+(2n-1)x^n$ is Arithmetico-Geometric sequence.
And we can find the sum to be $$S=x\left(\frac{1-[1+2(n-1)]x^n}{1-x}+\frac{2x(1-x^{n-1})}{(1-x)^2}\right)$$
However, I'm not able to simplify beyond this extent. The book says the function is continuous at $x=1$. So, the $(x-1)$ and $(1-x)$ must somehow not remain after simplifying. I can't see how I must proceed.
Also, I'm not allowed to use L'Hospital's rule or Taylor expansions. Please help me simplify this, or maybe suggest a new and better method.
Since $1+3+5+\cdots+(2n-1)=n^2$, you're just evaluating a derivative at $x=1$.