A series of natural numbers is divided into groups A series of natural numbers is divided into groups: $$(1),(2,3,4), (5,6,7,8,9,10,11), (12,13,......,26)$$ and so on.
The sum of numbers in $n^{th}$ group is $a.2^{2n-1}-(2n+b)2^{2n-1}+n+1$. Then find $a+b$
The main problem I am facing here is to find first term of $n^{th}$ group. Series for first term is like $1,2,5,12,27,...$ In this series I observed that difference of difference is in G.P. but don't know how to use this information.

The leading term of $n$-th group is $2^{n}-n$ so the last term is $2^{n+1}-n-2$.
Hence, the sum is 
\begin{align*}
  & \quad \frac{1}{2} (2^{n+1}-n-2+2^{n}-n)(2^{n+1}-n-2-2^{n}+n+1) \\
  &= (3\cdot 2^{n-1}-n-1)(2^{n}-1) \\
  &= 3\cdot 2^{2n-1}-3\cdot 2^{n-1}-n\cdot 2^{n}+n+1 \\
  &= 3\cdot 2^{2n-1}-(2n+3) 2^{n-1}+n+1
\end{align*}
