Bài 1.
$A=C_{2014}^1(3-1)+C_{2014}^2(3^2-1)+...+C_{2014}^{2014}(3^{2014}-1)$
$\Leftrightarrow A=[C_{2014}^1.3+C_{2014}^2.3^2+...+C_{2014}^{2014}.3^{2014}]
-[C_{2014}^1+C_{2014}^2+...+C_{2014}^{2014}]$
$\Leftrightarrow A=(4^{2014}-1)-(2^{2014}-1)=4^{2014}-2^{2014}$
Bài 2.
$C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n}=2^{20}-1$
$\Leftrightarrow C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n}+
C_{2n+1}^{2n+1} =2^{20}$
Áp dụng công thức $C_{n}^k=C_n^{n-k}$
Như vậy ta có :
$C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^n$
$=> C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1} = \dfrac{1}{2}[C_{2n+1}^0+C_{2n+1}^1
+...+C_{2n+1}^{2n+1}]$
$<=> C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1} = 2^{2n}$
$<=> 2^{2n}=2^{20}$
$<=> n=10$