I've been stuck on this problem for 5 days trying to figure out how to simplify the sum of this series.
The series is: ${100C0}*\frac{1}{3}^{100}*6^{0} + {100C_1}*\frac{1}{3}^{101}*6^{1} + {100C_2}*\frac{1}{3}^{102}*6^{2} +...+ {100C_{100}}*\frac{1}{3}^{200}*6^{100}$
To which I noticed the series is equivalent to $$ \sum_{i=0}^{100} {100C_i}\frac{1}{3}^{100+i}*6^i $$ From here I factored out $\frac{1}{3}^{100}$ giving me $ \frac{1}{3}^{100}*\sum_{i=0}^{100} {100C_i}*\frac{1}{3}^{i}*6^i $
Next I found that $\frac{1}{3}^{i}*6^i= 2^i$. So now I have deduced that the series is = to $$ \frac{1}{3}^{100}*\sum_{i=0}^{100} {100C_i}*2^i $$
However for the life of me I have no idea what to do next and am unsure if this is even correct. I don't know how to further simplify the sum.
Any help would be most appreciated.
$$S= \sum_{k=0}^{100} \binom{100}{k} (\frac{1}{3})^{100+k} 6^k=\frac{1}{3^{100}} \sum_{k=0}^{100} \binom{100}{k} 2^k $$
Now, consider the binomial expansion :
$$ (1+x)^{100} = \sum_{k=0}^{100} \binom{100}{k} x^k$$
Can you finish now?