1+5+13+29+61+.............n terms then find sum to n terms

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$$\begin{gathered} \mathrm{S}=1+5+13+29+61+........+{\mathrm{T}}_{\mathrm{n}-1}+{\mathrm{T}}_{\mathrm{n}} \\ \mathrm{S}= 1+5+13+29+61+........+{\mathrm{T}}_{\mathrm{n}-1}+{\mathrm{T}}_{\mathrm{n}} \\ \mathrm{Subtracting} \mathrm{the} \mathrm{above} \mathrm{two} \mathrm{equations} \mathrm{we} \mathrm{get}, \\ 0=1+4+8+16+32+.........+ \left( {\mathrm{T}}_{\mathrm{n}}-{\mathrm{T}}_{\mathrm{n}-1}\right)-{\mathrm{T}}_{\mathrm{n}} \\ \Rightarrow {\mathrm{T}}_{\mathrm{n}}=1+\underset{\mathrm{n}-1 \mathrm{terms}}{\underbrace{4+8+16+32+.........+ \left( {\mathrm{T}}_{\mathrm{n}}-{\mathrm{T}}_{\mathrm{n}-1}\right)}} \\ 4+8+16+32+......... \mathrm{is} \mathrm{a} \mathrm{G}.\mathrm{P}. \mathrm{of} \mathrm{n}-1 \mathrm{terms} \mathrm{having} \mathrm{first} \mathrm{term}\left(\mathrm{a}\right)=2 \mathrm{and} \\ \mathrm{common} \mathrm{ratio}\left(\mathrm{r}\right)=2 \\ \mathrm{Sum}=\frac{4\left(2^{\mathrm{n}-1}-1\right)}{2-1} \\ =2^2\left(2^{\mathrm{n}-1}-1\right) \\ =2^2.2^{\mathrm{n}-1}-2^2 \\ =2^{\mathrm{n}-1+2}-4=2^{\mathrm{n}+1}-4 \\ \Rightarrow {\mathrm{T}}_{\mathrm{n}}=1+2^{\mathrm{n}+1}-4 \\ \Rightarrow {\mathrm{T}}_{\mathrm{n}}=2^{\mathrm{n}+1}-3 \\ {\mathrm{S}}_{\mathrm{n}}={\sum }_{\mathrm{n}=1}^{\mathrm{n}=\mathrm{n}} \left({\mathrm{T}}_{\mathrm{n}}\right) \\ ={\sum }_{\mathrm{n}=1}^{\mathrm{n}=\mathrm{n}} \left(2^{\mathrm{n}+1}-3\right) \\ ={\sum }_{\mathrm{n}=1}^{\mathrm{n}=\mathrm{n}} \left(2^{\mathrm{n}+1}\right)-3{\sum }_{\mathrm{n}=1}^{\mathrm{n}=\mathrm{n}}\left(1\right) \\ =2^2+2^3+2^4+......+2^{\mathrm{n}+1}-3\left(1+1+1+1+....\mathrm{n} \mathrm{times}\right) \\ =\frac{2^2\left(2^{\mathrm{n}}-1\right)}{2-1}-3\mathrm{n} \\ =2^2.2^{\mathrm{n}}-2^2-3\mathrm{n} \\ {\mathrm{S}}_{\mathrm{n}}=2^{\mathrm{n}+2}-4-3\mathrm{n} \\ \mathbf{Hence}\mathbf{ }\mathbf{sum}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{S}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{5}\mathbf{+}\mathbf{13}\mathbf{+}\mathbf{29}\mathbf{+}\mathbf{61}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{ }\mathbf{upto}\mathbf{ }\mathbf{n}\mathbf{ }\mathbf{terms}\mathbf{ }\mathbf{is}\mathbf{ }{\mathbf{2}}^{\mathbf{n}\mathbf{+}\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{n}\mathbf{-}\mathbf{4}\mathbf{.} \end{gathered}$$

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