Chủ đề này có 3 trả lời

[thang96]

Tính tổng S=$C_{2013}^{1}\textrm{}+C_{2013}^{5}\textrm{}+C_{2013}^{9}\textrm{}+...+C_{2013}^{2009}\textrm{}+C_{2013}^{2013}\textrm{}$

[hxthanh]

$(1+1)^{4n+1}=C_{4n+1}^0+C_{4n+1}^1+C_{4n+1}^2+C_{4n+1}^3+C_{4n+1}^4+...+C_{4n+1}^{4n+1}\quad(1)$

$(1-1)^{4n+1}=C_{4n+1}^0-C_{4n+1}^1+C_{4n+1}^2-C_{4n+1}^3+C_{4n+1}^4-...-C_{4n+1}^{4n+1}\quad(2)$

$(1+i)^{4n+1}=C_{4n+1}^0+iC_{4n+1}^1-C_{4n+1}^2-iC_{4n+1}^3+C_{4n+1}^4+...+iC_{4n+1}^{4n+1}\quad(3)$

$(1-i)^{4n+1}=C_{4n+1}^0-iC_{4n+1}^1-C_{4n+1}^2+iC_{4n+1}^3+C_{4n+1}^4-...-iC_{4n+1}^{4n+1}\quad(4)$

 

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 $S_{503}=2^{2011}-2^{1005}$

 

[hxthanh]

Tổng quát hơn một chút ta có:

$C_n^1+C_n^5+...+C_n^{4k+1}=2^{n-2}+(\sqrt 2)^{n-2}\sin\frac{n\pi}{4}$

với $k=\left\lfloor\dfrac{n-1}{4}\right\rfloor$

 

Còn lại dành cho bạn:

a) $C_n^0+C_n^4+...+C_n^{4k}=?\quad$ với $k=\left\lfloor \dfrac{n}{4}\right\rfloor$

b) $C_n^2+C_n^6+...+C_n^{4k+2}=?\quad$ với $k=\left\lfloor \dfrac{n-2}{4}\right\rfloor$

c) $C_n^3+C_n^7+...+C_n^{4k+3}=?\quad$ với $k=\left\lfloor \dfrac{n-3}{4}\right\rfloor$

[thang96]

$(1+1)^{4n+1}=C_{4n+1}^0+C_{4n+1}^1+C_{4n+1}^2+C_{4n+1}^3+C_{4n+1}^4+...+C_{4n+1}^{4n+1}\quad(1)$

$(1-1)^{4n+1}=C_{4n+1}^0-C_{4n+1}^1+C_{4n+1}^2-C_{4n+1}^3+C_{4n+1}^4-...-C_{4n+1}^{4n+1}\quad(2)$

$(1+i)^{4n+1}=C_{4n+1}^0+iC_{4n+1}^1-C_{4n+1}^2-iC_{4n+1}^3+C_{4n+1}^4+...+iC_{4n+1}^{4n+1}\quad(3)$

$(1-i)^{4n+1}=C_{4n+1}^0-iC_{4n+1}^1-C_{4n+1}^2+iC_{4n+1}^3+C_{4n+1}^4-...-iC_{4n+1}^{4n+1}\quad(4)$

 

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