The sum of first $n$ terms of the sequence $1,1+2\left(1+2+2^2\right),\cdots$ $\left(1+2+2^2+\cdots +2^{k-1}\right)$ is of the form $2^{n+R}+S{n}^2+Tn+U,\mathrm{\forall }n\in N$. The value of $(R+S+T+U)$ is

Given sequence,  $1,1+2,\left(1+2+2^2\right)\cdot \cdots \left(1+2+2^2+\cdots +2^{k-1}\right)$

$\therefore$ The nth term of the above sequence is 

$T_n=1+2+2^2+\cdots +2^{n-1}$

$\begin{array}{l}=\frac{1\left(2^n-1\right)}{2-1}\left[\because a+ar+a{r}^2+\cdots +a^{n-1}=\frac{a\left(r^n-1\right)}{r-1},r>1\right]\\ =\frac{2^n-1}{1}\\ =2^n-1\end{array}$

Sum of $n$ terms of the series, 

$\begin{array}{l}\mathrm{\Sigma }{T}_n=\mathrm{\Sigma }\left(2^n-1\right)\\ =\left[2+2^2+2^3+2^4+\cdots +n\text{ terms }\right]\\ =\frac{2\left(2^n-1\right)}{(2-1)}-n\end{array}$

But it is given that, sum of $n$ terms of the series is

$\begin{array}{l}{2}^{n+R}+S\cdot n^2+T\cdot n+U.\\ \therefore 2^{n+R}+S\cdot n^2+T\cdot n+U=2^{n+1}-n-2\\ \Rightarrow R=1,S=0,T=-1,U=-2\\ \therefore R+S+T+U=1+0-1-2=-2\end{array}$