How do I add the terms in the binomial expansion of $(100+2)^6$? So, I stumbled upon the following question.
Using binomial theorem compute $102^6$.
Now, I broke the number into 100+2.
Then, applying binomial theorem
$\binom {6} {0}$$100^6(1)$+$\binom {6} {1}$$100^5(2)$+....

I stumbled upon this step. How did they add the humongous numbers? I am really confused.
Kindly help me clear my query.

The long way:
$$
\begin{align}
& 10^{12} + 12 \cdot 10^{10} + 6 \cdot 10^9 + 16 \cdot 10^7 + 24 \cdot 10^5 +192 \cdot 10^2 + 64 \\
=\; & 10^{12} + (10+2) 10^{10} + 6 \cdot 10^9 + (10+6) 10^7 + (20+4) 10^5 +(100+90 +2) 10^2 + 60+4 \\
=\; & \color{red}1\cdot10^{12} + \color{red}1 \cdot 10^{11} + \color{red}2\cdot 10^{10} + \color{red}6 \cdot 10^9 + \color{red}1 \cdot 10^8 + \color{red}6 \cdot 10^7+\color{red}2 \cdot 10^6 + \color{red}4 \cdot 10^5+\color{red}1 \cdot 10^4+\color{red}9\cdot 10^3 + \color{red}2 \cdot 10^2 + \color{red}6 \cdot 10^1 + \color{red}4 \cdot 10^0
\end{align}
$$
The latter is precisely the representation in base $\,10\,$ of $\;\color{red}{1126162419264}\,$.
