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.
 A: That number is $1\; 12 \; 6\; 16 \; 24\;192\; 64$ (space added for emphasis).  Notice a relationship with the coefficients of the powers of $10$?
A: 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 $\;\color{red}{1126162419264}\,$.
A: $10^{12} + 12\times10^{10} + 6\times10^9 + 16*10^7 + 24\times10^5 + 192\times10^2 + 64$
$= 10^{12} + 10^{11} + 2\times10^{10} + 6\times10^9 + 10^8 + 6\times10^7 + 2\times10^6 + 4\times10^5 + 10^4 + 9\times10^3 + 2\times10^2 + 6\times10^1 + 4\times10^0= 1126162419264$
A: Since it's all powers of ten you could add those quite easily.
If it were something else you could have needed to use some calculator. But here you have just powers of tens, which basically keep everything the same. Here's the number with a few spaces to make you understand it.
$\text{1 12 6 16 24 192 64}$
Those powers just end up just putting those digits in the right order because we use a number system with ten digits.
A: Start adding from the rightmost term($64$); you will notice that the number of trailing zeros on the preceding term is exactly the same as the number of digits in following term.
$$
...2400000\quad +\\
.....19200\quad +\\
.......64\quad
$$
...and so on till the first.
