$(1+1)^{4n+1}=C_{4n+1}^0+C_{4n+1}^1+C_{4n+1}^2+C_{4n+1}^3+C_{4n+1}^4+...+C_{4n+1}^{4n+1}\quad(1)$
$(1-1)^{4n+1}=C_{4n+1}^0-C_{4n+1}^1+C_{4n+1}^2-C_{4n+1}^3+C_{4n+1}^4-...-C_{4n+1}^{4n+1}\quad(2)$
$(1+i)^{4n+1}=C_{4n+1}^0+iC_{4n+1}^1-C_{4n+1}^2-iC_{4n+1}^3+C_{4n+1}^4+...+iC_{4n+1}^{4n+1}\quad(3)$
$(1-i)^{4n+1}=C_{4n+1}^0-iC_{4n+1}^1-C_{4n+1}^2+iC_{4n+1}^3+C_{4n+1}^4-...-iC_{4n+1}^{4n+1}\quad(4)$
Lấy $(1) - (2)$, cho ta:
$\quad 2^{4n+1}=2(C_{4n+1}^1+C_{4n+1}^3+C_{4n+1}^5+...+C_{4n+1}^{4n+1})$
$\Leftrightarrow 2^{4n}=C_{4n+1}^1+C_{4n+1}^3+C_{4n+1}^5+...+C_{4n+1}^{4n+1}\quad(5)$
Lấy $(3) - (4)$, cho ta:
$\quad(1+i)^{4n+1}-(1-i)^{4n+1}=2i(C_{4n+1}^1-C_{4n+1}^3+C_{4n+1}^5-...+C_{4n+1}^{4n+1})$
$\Leftrightarrow \dfrac{(1+i)^{4n+1}-(1-i)^{4n+1}}{2i}=C_{4n+1}^1-C_{4n+1}^3+C_{4n+1}^5-...+C_{4n+1}^{4n+1}\quad(6)$
Cuối cùng, lấy $(5)+(6)$ ta được:
$2^{4n}+\dfrac{(1+i)^{4n+1}-(1-i)^{4n+1}}{2i}=2(C_{4n+1}^1+C_{4n+1}^5+...+C_{4n+1}^{4n+1})$
hay
$\boxed{\displaystyle S_n=C_{4n+1}^1+C_{4n+1}^5+...+C_{4n+1}^{4n+1}=2^{4n-1}+\dfrac{(1+i)^{4n+1}-(1-i)^{4n+1}}{4i}}\quad(*)$
Bây giờ ta sẽ rút gọn vế phải của $(*)$, ta có:
$(1+i)^{4n+1}=\left[\sqrt 2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\right]^{4n+1}=2^{2n}\sqrt 2\left(\cos\frac{\pi(4n+1)}{4}+i\sin\frac{\pi(4n+1)}{4}\right)$
$(1-i)^{4n+1}=\left[\sqrt 2\left(\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}\right)\right]^{4n+1}=2^{2n}\sqrt 2\left(\cos\frac{\pi(4n+1)}{4}-i\sin\frac{\pi(4n+1)}{4}\right)$
Do đó:
$\dfrac{(1+i)^{4n+1}-(1-i)^{4n+1}}{4i}=2^{2n-1}\sqrt 2\sin\frac{\pi(4n+1)}{4}=2^{2n-1}\sqrt 2\cdot \frac{(-1)^n\sqrt 2}{2}=(-1)^n 2^{2n-1}$
Kết quả cuối cùng ta được:
$S_n=2^{4n-1}+(-1)^n2^{2n-1}$
Bài toán của bạn, chỉ cần thay $n=503$ vào...
Kết quả là $S_{503}=2^{2011}-2^{1005}$
thế i là gì vậy bạn