Is the number :
${100 \choose 0}\cdot2^0 + {100 \choose 1}\cdot2^1+{100 \choose 2}\cdot2^2+{100 \choose 3}\cdot2^3 \ldots + {100 \choose 100}\cdot2^{100}$
divisible by 3?
I tried looking at the first few terms which can be written as :
$2^0 + \frac{100}{1}\cdot 2^1 + \frac{100 \cdot 99}{2!}\cdot2^2+\frac{100 \cdot 99\cdot98}{3!}\cdot2^3+\ldots$
So I my guess it that the general term
$a_n = \frac{\prod_{i=0}^n 100-i}{n!}\cdot2^n$
but I am not sure how to continue from here to obtain the final value and check if it is divisible by 3. Is there another way?
Hint: write the binomial expansion of $${(1+2)}^{100}$$ Do you see a similarity?