Intermediate sum

There's no need to use anything fancier than a scientific calculator. Since we only want a limited precision, let's just use the last 8 terms (since the rest are insignificant) and write them as

$(4+4^{2}+4^{3}+4^{4}+4^{5}+4^{6}+4^{7}+4^{8})\cdot +4^{2010}$

$=87380\cdot 4^{2010}$

The take a common logarithm to help put in scientific notation

$\log {(87380\cdot 4^{2010})}=\log {87380} + 2010 \log{4} = 1215.81995$

The division just strips off the whole number part: $1215.81995-1215=0.81995$

Leaving the answer as $10^{0.81995}=\boxed{1.2077988}$

Let $S=4+4^2+4^3+4^4+4^5+4^6+\dots+4^{2018}$

$\begin{array}{cccc} S=4+4^2+4^3+4^4+4^5+4^6+\dots+4^{2018} \\ -4S=4^2+4^3+4^4+4^5+4^6+\dots+4^{2019}\\ \hline -3S=4-4^{2019} \end{array}$

$-3S=4-4^{2019} \implies 3S=4^{2019}-4 \implies S=\frac{4^{2019}-4}{3} \approx 1.2078 \times 10^{1215}$

So $\frac{1.2078 \times 10^{1215}}{10^{1215}}=\boxed{\large{1.2078}}$

Note

• For every integer $a \ne 1$ of $S=a+a^2+a^3+a^4+a^5+a^6+\dots+a^n$ that $S=\frac{a^{n+1}-a}{a-1}$