Ứng dụng số phức giải toán khai triển, tính tổng nhị thức Niutơn

Phương pháp
Ta nhắc lại công thức khai triển nhị thức Niutơn:
${\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {C_n^k} {a^{n – k}}{b^k}$ $ = C_n^o{a^n} + C_n^1{a^{n – 1}}b + C_n^1{a^{n – 2}}{b^2}$ $ + … + C_n^{n – 1}a{b^{n – 1}} + C_n^n{b^n}.$
Ta lưu ý rằng $\forall m \in {N^*}$ thì ${i^{4m}} = 1$, ${i^{4m + 1}} = i$, ${i^{4m + 2}} = – 1$, ${i^{4m + 3}} = – i.$

Các ví dụ điển hình thường gặp
Ví dụ 1. Tính tổng:
a. ${S_1} = 1 – C_n^2 + C_n^4 – C_n^6 + … .$

b. ${S_2} = C_n^1 – C_n^3 + C_n^5 – C_n^7 + … .$

Ta có:

${\left( {1 + i} \right)^n}$ $ = 1 + C_n^1i + C_n^2{i^2} + … + C_n^n{i^n}$

$ = \left( {1 – C_n^2 + C_n^4 – C_n^6 + …} \right)$ $ + i\left( {C_n^1 – C_n^3 + C_n^5 – C_n^7 + …} \right) (1).$

${\left( {1 + i} \right)^n}$ $ = \sqrt {{2^n}} c{\rm{os}}\frac{{n\pi }}{4} + i\sqrt {{2^n}} {\rm{sin}}\frac{{n\pi }}{4} (2).$

Từ $(1)$ và $(2)$ suy ra:

${{\rm{S}}_1} = \sqrt {{2^n}} c{\rm{os}}\frac{{n\pi }}{4}.$

${S_2} = \sqrt {{2^n}} {\rm{sin}}\frac{{n\pi }}{4}.$

Ví dụ 2. Chứng minh rằng: $C_{100}^0 – C_{100}^2 + C_{100}^4 – C_{100}^6$ $ + … – C_{100}^{98} + C_{100}^{100} = – {2^{50}}.$

${\left( {1 + i} \right)^{100}}$ $ = C_{100}^0 + C_{100}^1i + C_{100}^2{i^2} + … + C_{100}^{100}{i^{100}}$

$ = \left( {C_{100}^0 – C_{100}^2 + C_{100}^4 – … + C_{100}^{100}} \right)$ $ + \left( {C_{100}^1 – C_{100}^3 + C_{100}^5 + … – C_{100}^{99}} \right)i.$

${\left( {1 + i} \right)^2} = 2i$ $ \Rightarrow {\left( {1 + i} \right)^{100}} = {\left( {2i} \right)^{50}} = – {2^{50}}.$
Vậy: $C_{100}^0 – C_{100}^2 + C_{100}^4 – … + C_{100}^{100} = – {2^{50}}.$

Ví dụ 3. Tính các tổng sau:

$A = C_{15}^0 – 3C_{15}^2 + 5C_{15}^4 – 7C_{15}^6$ $ + …. + 13C_{15}^{12} – 15C_{15}^{14}.$

$B = 2C_{15}^1 – 4C_{15}^3 + 6C_{15}^5 – 8C_{15}^7$ $ + …. + 14C_{15}^{13} – 16C_{15}^{15}.$

Xét khai triển:
${\left( {1 + x} \right)^{15}}$ $ = C_{15}^0 + C_{15}^1x + C_{15}^2{x^2} + C_{15}^3{x^3}$ $ + … + C_{15}^{12}{x^{12}} + C_{15}^{13}{x^{13}} + C_{15}^{14}{x^{14}} + C_{15}^{15}{x^{15}}$
$ \Rightarrow x{\left( {1 + x} \right)^{15}}$ $ = C_{15}^0x + C_{15}^1{x^2} + C_{15}^2{x^3} + C_{15}^3{x^4}$ $ + … + C_{15}^{12}{x^{13}} + C_{15}^{13}{x^{14}} + C_{15}^{14}{x^{15}} + C_{15}^{15}{x^{16}}.$
Lấy đạo hàm hai vế:
${\left( {1 + x} \right)^{15}} + 15x{\left( {1 + x} \right)^{14}}$

$ = C_{15}^0 + 2C_{15}^1x + 3C_{15}^2{x^2} + 4C_{15}^3{x^3}$ $ + … + 13C_{15}^{12}{x^{12}} + 14C_{15}^{13}{x^{13}}$ $ + 15C_{15}^{14}{x^{14}} + 16C_{15}^{15}{x^{15}}.$
Thay $x$ bởi $i$ ta được:

${\left( {1 + i} \right)^{15}} + 15i{\left( {1 + i} \right)^{14}}$ $ = C_{15}^0 + 2C_{15}^1i + 3C_{15}^2{i^2} + 4C_{15}^3{i^3}$ $ + … + 13C_{15}^{12}{i^{12}} + 14C_{15}^{13}{i^{13}}$ $ + 15C_{15}^{14}{i^{14}} + 16C_{15}^{15}{i^{15}}$
= (${C_{15}^0 – 3C_{15}^2 + 5C_{15}^4 – 7C_{15}^6}$ ${ + …. + 13C_{15}^{12} – 15C_{15}^{14}}$) + (${2C_{15}^1 – 4C_{15}^3 + 6C_{15}^5 – 8C_{15}^7}$ ${ + …. + 14C_{15}^{13} – 16C_{15}^{15}}$)$i.$

Mặt khác:

${\left( {1 + i} \right)^{15}} + 15i{\left( {1 + i} \right)^{14}}$ $ = \sqrt {{2^{15}}} {\left( {c{\rm{os}}\frac{\pi }{4} + {\rm{i}}\sin \frac{\pi }{4}} \right)^{15}}$ $ + 15i\sqrt {{2^{14}}} {\left( {c{\rm{os}}\frac{\pi }{4} + {\rm{i}}\sin \frac{\pi }{4}} \right)^{14}}$

$ = \sqrt {{2^{15}}} \left( {\frac{{\sqrt 2 }}{2} – \frac{{\sqrt 2 }}{2}i} \right) + 15i{.2^7}\left( { – i} \right)$ $ = {2^7} – {2^7}i + {15.2^7}$ $ = {16.2^7} – {2^7}i = {2^{11}} – {2^7}i.$
Vậy:
$A = C_{15}^0 – 3C_{15}^2 + 5C_{15}^4 – 7C_{15}^6$ $ + …. + 13C_{15}^{12} – 15C_{15}^{14} = {2^{11}}.$
$B = 2C_{15}^1 – 4C_{15}^3 + 6C_{15}^5 – 8C_{15}^7$ $ + …. + 14C_{15}^{13} – 16C_{15}^{15} = – {2^7}.$

Ví dụ 4. Chứng minh rằng:

${S_1} = C_n^0 – C_n^2 + C_n^4 – C_n^6 + C_n^8 – …$ $ = {\left( {\sqrt 2 } \right)^n}\cos \frac{{n\pi }}{4}.$

${S_2} = C_n^1 – C_n^3 + C_n^5 – C_n^7 + C_n^9 – …$ $ = {\left( {\sqrt 2 } \right)^n}\sin \frac{{n\pi }}{4}.$

Xét khai triển nhị thức Newton:

${\left( {1 + i} \right)^n}$ $ = C_n^0 + iC_n^1 + {i^2}C_n^2 + {i^3}C_n^3 + {i^4}C_n^4$ $ + … + {i^{n – 1}}C_n^{n – 1} + {i^n}C_n^n.$

Vì ${i^k} = \left\{ \begin{array}{l}
1, (k = 4m)\\
i, (k = 4m + 1)\\
– 1, (k = 4m + 2)\\
– i, (k = 4m + 3)
\end{array} \right.$ với $m \in {{\rm Z}^ + }$, nên ta có:

${\left( {1 + i} \right)^n}$ $ = C_n^0 – C_n^2 + C_n^4 – …$ $ + i\left( {C_n^1 – C_n^3 + C_n^5 – ….} \right).$

Mặt khác, theo công thức Moivre thì:

${\left( {1 + i} \right)^n}$ $ = {\left( {\sqrt 2 } \right)^n}{\left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)^n}$ $ = {\left( {\sqrt 2 } \right)^n}\left( {\cos \frac{{n\pi }}{4} + i\sin \frac{{n\pi }}{4}} \right).$

Từ $(1)$ và $(2)$ ta có điều phải chứng minh.

Ví dụ 5. Tính tổng $S = \frac{1}{2}C_{2n}^1 – \frac{1}{4}C_{2n}^3 + \frac{1}{6}C_{2n}^5 – \frac{1}{8}C_{2n}^7 + …$

Chú ý rằng $\frac{1}{{2k}}C_{2n}^{2k – 1} = \frac{1}{{2n + 1}}C_{2n + 1}^{2k}$ nên:

$S = \frac{1}{2}C_{2n}^1 – \frac{1}{4}C_{2n}^3 + \frac{1}{6}C_{2n}^5 – \frac{1}{8}C_{2n}^7 + …$
$ = \frac{1}{{2n + 1}}C_{2n + 1}^2 – \frac{1}{{2n + 1}}C_{2n + 1}^4$ $ + \frac{1}{{2n + 1}}C_{2n + 1}^6 – \frac{1}{{2n + 1}}C_{2n + 1}^8 + …$

$ = \frac{1}{{2n + 1}}$.$\left( {C_{2n + 1}^2 – C_{2n + 1}^4 + C_{2n + 1}^6 – C_{2n + 1}^8 + …} \right).$

Vì ${\left( {1 + i} \right)^{2n + 1}}$ $ = \left( {C_{2n + 1}^0 – C_{2n + 1}^2 + C_{2n + 1}^4 – …} \right)$ $ + i\left( {C_{2n + 1}^1 – C_{2n + 1}^3 + C_{2n + 1}^5 – …} \right).$

Và ${\left( {1 + i} \right)^{2n + 1}}$ $ = {\left( {\sqrt 2 } \right)^{2n + 1}}$ $\left( {\cos \frac{{2n + 1}}{4}\pi + i\sin \frac{{2n + 1}}{4}\pi } \right)$ nên:

$C_{2n + 1}^0 – C_{2n + 1}^2 + C_{2n + 1}^4 – C_{2n + 1}^6$ $ + … = {\left( {\sqrt 2 } \right)^{2n + 1}}\cos \frac{{2n + 1}}{4}\pi .$

Vậy ta có $S = \frac{1}{{2n + 1}}$ $\left[ {1 – {{\left( {\sqrt 2 } \right)}^{2n + 1}}\cos \frac{{2n + 1}}{4}\pi } \right].$

Ví dụ 6. Tính tổng: $(n \in {{\rm Z}^ + }).$

$A = C_n^0\cos a + C_n^1\cos 2a + C_n^2\cos 3a$ $ + … + C_n^{n – 1}\cos na + C_n^n\cos (n + 1)a.$

$B = C_n^0\sin a + C_n^1\sin 2a + C_n^2\sin 3a$ $ + … + C_n^{n – 1}\sin na + C_n^n\sin (n + 1)a.$

Đặt $z = \cos a + i\sin a$ thì ${z^n} = \cos na + i\sin na.$

Do đó ta có:

$A + iB = C_n^0\left( {\cos a + i\sin a} \right)$ $ + C_n^1\left( {\cos 2a + i\sin 2a} \right)$ $ + C_n^2\left( {\cos 3a + i\sin 3a} \right)$

$ + … + C_n^{n – 1}\left( {\cos na + i\sin na} \right)$ $ + C_n^n\left( {\cos (n + 1)a + i\sin (n + 1)a} \right)$

$ = z\left( {C_n^0 + C_n^1z + C_n^2{z^2} + C_n^3{z^3} + … + C_n^n{z^n}} \right)$ $ = z{\left( {1 + z} \right)^n}.$

Vì $1 + z = 1 + \cos a + i\sin a$ $ = 2\cos \frac{a}{2}\left( {\cos \frac{a}{2} + i\sin \frac{a}{2}} \right).$

Nên: $A + iB = \left( {\cos a + i\sin a} \right)$.${\left[ {2\cos \frac{a}{2}\left( {\cos \frac{a}{2} = i\sin \frac{a}{2}} \right)} \right]^n}$

$ = {2^n}{\cos ^n}\frac{a}{2}\left( {\cos a + i\sin a} \right)$.$\left( {\cos \frac{{na}}{2} + i\sin \frac{{na}}{2}} \right)$

$ = {2^n}{\cos ^n}\frac{a}{2}$.$\left( {\cos \frac{{n + 2}}{2}a + i\sin \frac{{n + 2}}{2}a} \right)$

Vậy $A = {2^n}{\cos ^n}\frac{a}{2}\cos \frac{{n + 2}}{2}a$, $B = {2^n}{\cos ^n}\frac{a}{2}\sin \frac{{n + 2}}{2}a.$

Nhận xét: Cho $n$ là giá trị cụ thể, suy ra được nhiều biểu thức lượng giác đẹp.
Ví dụ: $\cos a + 5\cos 2a + 10\cos 3a$ $ + 10\cos 4a + 5\cos 5a + \cos 6a$ $ = {2^5}{\cos ^5}\frac{a}{2}\cos \frac{{7a}}{2}.$

Author: admin