In the sequence $1,4,11,26$… each term is $2⋅(n-1)^{th}$ term $+ n$. What is the $n^{th}$ term? I readily see that it is $2^{n+1} - (n+2)$ but how can I deduce the $n^{th}$ term from the given pattern i.e. $2⋅(n−1)^{th}$ term $+n$ .

So we have $a_0=0$ and $a_n=2a_{n-1}+n$ for $n\ge 1$. Let $A(x)=\sum_{n=0}^{\infty}{a_n x^n}$. Then multiplying both sides of the recurrence by $x^n$ and summing over $n\ge 1$, we get
$$
\begin{split}
A(x)=A(x)-a_0&=2xA(x)+\sum_{n\ge 1}{n x^n}\\
&=2xA(x)+x\left(\sum_{n\ge 0}{x^n}\right)'\\
&=2xA(x)+x\left(\frac{1}{1-x}\right)'\\
&=2xA(x)+\frac{x}{(1-x)^2},
\end{split}
$$
so $(1-2x)A(x)=\dfrac{x}{(1-x)^2}$ and hence
$$
A(x)=\frac{x}{(1-2x)(1-x)^2}=\dfrac{2}{1-2x}-\dfrac{1}{(1-x)^2}-\frac{1}{1-x},
$$
so that $a_n=[x^n]A(x)=2\cdot2^n-(n+1)-1=2^{n+1}-n-2$.
