This is exponential?

$\large \displaystyle {\color{#20A900} 1+6(1)}+6(7)+6(7^{2})+....+6(7^{99})$

$\large \displaystyle {\color{#20A900}7+6(7) }+6(7^{2})+....+6(7^{99})$

$\large \displaystyle {\color{#20A900}7^{2}+6(7^{2}) }+....+6(7^{99})$

.....

$\large \displaystyle {\color{#20A900}7^{99}+6(7^{99})}$

$\large \displaystyle {\color{#D61F06}7^{100}}$

$\large \displaystyle {\color{#20A900}e^{100ln7}}$

$\begin{aligned} e^\alpha & = 1 + 6(1) + 6(7) + 6(7^2) + 6(7^3) + \cdots + 6(7^{99}) \\ \frac {e^\alpha - 1}6 & = 7^0 + 7^1 + 7^2 + 7^3 + \cdots + 7^{99} \\ & = \frac {7^{100}-1}{7-1} \\ \implies \frac {{\color{#3D99F6}e^\alpha} - 1}6 & = \frac {{\color{#3D99F6}7^{100}}-1}6 \\ \implies e^\alpha & = 7^{100} \\ \implies \alpha & = 100 \ln 7 \approx \boxed{194.59} \end{aligned}$

We can rewrite the expression as:

$1+6(1+7+7^{2}+ 7^{3}+....7^{99})$ = $e^{\alpha}$

it's easy to observe it forms GP.

The sum can be found using the formula ${ S }_{ n }=\frac { a({ r }^{ n }-1) }{ r-1 }$

$1+6(\frac { { 7 }^{ 100 }-1 }{ 7-1 })$ = $e^{\alpha}$ .

Which is = $1+7^{100} -1= e^{\alpha}$

Now finally we get $(7^{100})= e^{\alpha}$ .

Take natural log on both sides

Finally we would get $\alpha =100ln7$