geometric series calculating the sum for first 45 terms I am able to calculate up to the step in blue, but I cant understand how do I simply the terms from -3^45 to 10^20 ?
Can anyone explain?

 A: $$S_{45}=-\frac{1}{8}(1-(-3)^{45})=-\frac{1}{8}+\frac{1}{8}(-3)^{45}$$
$$=-\frac{1}{8}-\frac{1}{8}(3)^{45}=-\frac{1}{8}-\frac{1}{8}(3^5)^9$$
Since $ $ $3^5=3\times3\times3\times3\times3=243$, we have
$$-\frac{1}{8}-\frac{1}{8}(243)^9=-\frac{1}{8}-\frac{1}{8}(2.43\times10^2)^9=-\frac{1}{8}-\frac{1}{8}(2.43^9\times10^{18})$$
Since $ $ $2.43^9=2954.31270…$  (you can either do this manually by multiplying 2.43 by 2.43, systematically, 9 times (not recommended), or use a calculator), we have
$$-\frac{1}{8}-\frac{1}{8}(2954.31270…\times10^{18})$$
$$=-\frac{1}{8}-\frac{2954.31270…}{8}\times10^{18}$$
$$=-\frac{1}{8}-369.2890…\times10^{18}$$
$$=-\frac{1}{8}-3.692890…\times10^2\times10^{18}$$
$$=-\frac{1}{8}-3.692890…\times10^{20}$$
$$\simeq-3.69\times10^{20}$$
A: $S_{45} = -\frac{3^{45}}{8}-\frac{1}{8}.$  Since the first fraction is so large, the second fraction can be ignored.  You can approximate the first fraction with log's.  
$$\ln -S_{45}=\ln\frac{3^{45}}{8} = 45\ln 3 -\ln 8 \approx  45(1.098)- 2.079 = 47.358. $$
Therefore $S_{45} \approx -e^{47.358} = -3.69\times 10^{20}$
It works even better with base-10 logs.
