Бином Ньютона для возведения в степень двучлена:
![(a+b)^n=C_n^0~a^n+C_n^1~a^{n-1}~b^1+...+C_n^{n-1}~a^1~b^{n-1}+C_n^n~b^n (a+b)^n=C_n^0~a^n+C_n^1~a^{n-1}~b^1+...+C_n^{n-1}~a^1~b^{n-1}+C_n^n~b^n](https://tex.z-dn.net/?f=%28a%2Bb%29%5En%3DC_n%5E0~a%5En%2BC_n%5E1~a%5E%7Bn-1%7D~b%5E1%2B...%2BC_n%5E%7Bn-1%7D~a%5E1~b%5E%7Bn-1%7D%2BC_n%5En~b%5En)
Формула сочетаний:
![C_n^m=\dfrac{n!}{m!(n-m)!} C_n^m=\dfrac{n!}{m!(n-m)!}](https://tex.z-dn.net/?f=C_n%5Em%3D%5Cdfrac%7Bn%21%7D%7Bm%21%28n-m%29%21%7D)
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1.
![C_6^0=\dfrac{6!}{0!\cdot(6-0)!}=\dfrac{6!}{1\cdot6!}=1;~~~~~~~~~C_6^6=C_6^1=1\\\\C_6^1=\dfrac{6!}{1!\cdot(6-1)!}=\dfrac{5!\cdot 6}{1\cdot5!}=6;~~~~~~~~~C_6^5=C_6^1=6\\\\C_6^2=\dfrac{6!}{2!\cdot(6-2)!}=\dfrac{4!\cdot 5\cdot 6}{1\cdot 2\cdot4!}=15;~~~~C_6^4=C_6^2=15\\\\C_6^3=\dfrac{6!}{3!\cdot(6-3)!}=\dfrac{3!\cdot 4\cdot 5\cdot 6}{1\cdot2\cdot 3\cdot3!}=20 C_6^0=\dfrac{6!}{0!\cdot(6-0)!}=\dfrac{6!}{1\cdot6!}=1;~~~~~~~~~C_6^6=C_6^1=1\\\\C_6^1=\dfrac{6!}{1!\cdot(6-1)!}=\dfrac{5!\cdot 6}{1\cdot5!}=6;~~~~~~~~~C_6^5=C_6^1=6\\\\C_6^2=\dfrac{6!}{2!\cdot(6-2)!}=\dfrac{4!\cdot 5\cdot 6}{1\cdot 2\cdot4!}=15;~~~~C_6^4=C_6^2=15\\\\C_6^3=\dfrac{6!}{3!\cdot(6-3)!}=\dfrac{3!\cdot 4\cdot 5\cdot 6}{1\cdot2\cdot 3\cdot3!}=20](https://tex.z-dn.net/?f=C_6%5E0%3D%5Cdfrac%7B6%21%7D%7B0%21%5Ccdot%286-0%29%21%7D%3D%5Cdfrac%7B6%21%7D%7B1%5Ccdot6%21%7D%3D1%3B~~~~~~~~~C_6%5E6%3DC_6%5E1%3D1%5C%5C%5C%5CC_6%5E1%3D%5Cdfrac%7B6%21%7D%7B1%21%5Ccdot%286-1%29%21%7D%3D%5Cdfrac%7B5%21%5Ccdot%206%7D%7B1%5Ccdot5%21%7D%3D6%3B~~~~~~~~~C_6%5E5%3DC_6%5E1%3D6%5C%5C%5C%5CC_6%5E2%3D%5Cdfrac%7B6%21%7D%7B2%21%5Ccdot%286-2%29%21%7D%3D%5Cdfrac%7B4%21%5Ccdot%205%5Ccdot%206%7D%7B1%5Ccdot%202%5Ccdot4%21%7D%3D15%3B~~~~C_6%5E4%3DC_6%5E2%3D15%5C%5C%5C%5CC_6%5E3%3D%5Cdfrac%7B6%21%7D%7B3%21%5Ccdot%286-3%29%21%7D%3D%5Cdfrac%7B3%21%5Ccdot%204%5Ccdot%205%5Ccdot%206%7D%7B1%5Ccdot2%5Ccdot%203%5Ccdot3%21%7D%3D20)
![\boldsymbol{(x-2y)^6=}\\\\=C_6^0~x^6+C_6^1~x^5~(-2y)^1+C_6^2~x^4~(-2y)^2+C_6^3~x^3~(-2y)^3+\\\\~~~~~+C_6^4~x^2~(-2y)^4+C_6^5~x^1~(-2y)^5+C_6^6~(-2y)^6=\\\\=1\cdot x^6-6x^5\cdot2y+15x^4\cdot4y^2-20x^3\cdot8y^3+\\\\~~~~~+15x^2\cdot 16y^4-6x\cdot32y^5+1\cdot 64y^6=\\\\=\boldsymbol{x^6-12x^5y+60x^4y^2-160x^3y^3+240x^2y^4-192xy^5+64y^6} \boldsymbol{(x-2y)^6=}\\\\=C_6^0~x^6+C_6^1~x^5~(-2y)^1+C_6^2~x^4~(-2y)^2+C_6^3~x^3~(-2y)^3+\\\\~~~~~+C_6^4~x^2~(-2y)^4+C_6^5~x^1~(-2y)^5+C_6^6~(-2y)^6=\\\\=1\cdot x^6-6x^5\cdot2y+15x^4\cdot4y^2-20x^3\cdot8y^3+\\\\~~~~~+15x^2\cdot 16y^4-6x\cdot32y^5+1\cdot 64y^6=\\\\=\boldsymbol{x^6-12x^5y+60x^4y^2-160x^3y^3+240x^2y^4-192xy^5+64y^6}](https://tex.z-dn.net/?f=%5Cboldsymbol%7B%28x-2y%29%5E6%3D%7D%5C%5C%5C%5C%3DC_6%5E0~x%5E6%2BC_6%5E1~x%5E5~%28-2y%29%5E1%2BC_6%5E2~x%5E4~%28-2y%29%5E2%2BC_6%5E3~x%5E3~%28-2y%29%5E3%2B%5C%5C%5C%5C~~~~~%2BC_6%5E4~x%5E2~%28-2y%29%5E4%2BC_6%5E5~x%5E1~%28-2y%29%5E5%2BC_6%5E6~%28-2y%29%5E6%3D%5C%5C%5C%5C%3D1%5Ccdot%20x%5E6-6x%5E5%5Ccdot2y%2B15x%5E4%5Ccdot4y%5E2-20x%5E3%5Ccdot8y%5E3%2B%5C%5C%5C%5C~~~~~%2B15x%5E2%5Ccdot%2016y%5E4-6x%5Ccdot32y%5E5%2B1%5Ccdot%2064y%5E6%3D%5C%5C%5C%5C%3D%5Cboldsymbol%7Bx%5E6-12x%5E5y%2B60x%5E4y%5E2-160x%5E3y%5E3%2B240x%5E2y%5E4-192xy%5E5%2B64y%5E6%7D)
2.
![(a+b)^n=256\\256=256^1=16^2=4^4=2^8=\left(\sqrt2\right)^{16}=\left(\sqrt[4]2\right)^{32}.... (a+b)^n=256\\256=256^1=16^2=4^4=2^8=\left(\sqrt2\right)^{16}=\left(\sqrt[4]2\right)^{32}....](https://tex.z-dn.net/?f=%28a%2Bb%29%5En%3D256%5C%5C256%3D256%5E1%3D16%5E2%3D4%5E4%3D2%5E8%3D%5Cleft%28%5Csqrt2%5Cright%29%5E%7B16%7D%3D%5Cleft%28%5Csqrt%5B4%5D2%5Cright%29%5E%7B32%7D....)
Не зная значение выражения в скобках, ответить однозначно на вопрос, чему равно n, невозможно.
Если
, то n , например, может быть равно 1, 2, 4, 8 и т.д.