![(1+x)^n=\Sigma_{k=0}^n (C_n^k*x^k)](https://tex.z-dn.net/?f=%281%2Bx%29%5En%3D%5CSigma_%7Bk%3D0%7D%5En+%28C_n%5Ek%2Ax%5Ek%29)
докажем методом математической индукции:
1) проверим для любого n. Пусть n=1
![(1+x)^1=\Sigma_{k=0}^1(C_1^k*x^k)=C_1^0*x^0+C_1^1*x^1=1+x](https://tex.z-dn.net/?f=%281%2Bx%29%5E1%3D%5CSigma_%7Bk%3D0%7D%5E1%28C_1%5Ek%2Ax%5Ek%29%3DC_1%5E0%2Ax%5E0%2BC_1%5E1%2Ax%5E1%3D1%2Bx)
2) пусть верно для n
докажем равенство для n+1
Для этого распишем данную сумму подробнее:
![(1+x)^n=(C_n^01+C_n^1*x^1+C_n^2*x^2+..+C_n^n*x^n)](https://tex.z-dn.net/?f=%281%2Bx%29%5En%3D%28C_n%5E01%2BC_n%5E1%2Ax%5E1%2BC_n%5E2%2Ax%5E2%2B..%2BC_n%5En%2Ax%5En%29)
запишем эту сумму для n+1
![(1+x)^{n+1}=(1+x)*(1+x)^n=](https://tex.z-dn.net/?f=%281%2Bx%29%5E%7Bn%2B1%7D%3D%281%2Bx%29%2A%281%2Bx%29%5En%3D)
![=(1+x)*(C_n^01+C_n^1*x^1+C_n^2*x^2+..+C_n^n*x^n)=](https://tex.z-dn.net/?f=%3D%281%2Bx%29%2A%28C_n%5E01%2BC_n%5E1%2Ax%5E1%2BC_n%5E2%2Ax%5E2%2B..%2BC_n%5En%2Ax%5En%29%3D)
раскроем скобки
![=(C_n^01+C_n^1*x^1+C_n^2*x^2+..+C_n^n*x^n)+ +x*((C_n^01+C_n^1*x^1+C_n^2*x^2+..+C_n^n*x^n)) ](https://tex.z-dn.net/?f=%3D%28C_n%5E01%2BC_n%5E1%2Ax%5E1%2BC_n%5E2%2Ax%5E2%2B..%2BC_n%5En%2Ax%5En%29%2B+%0A%0A%2Bx%2A%28%28C_n%5E01%2BC_n%5E1%2Ax%5E1%2BC_n%5E2%2Ax%5E2%2B..%2BC_n%5En%2Ax%5En%29%29%0A%0A)
![(C_n^01+C_n^1*x^1+C_n^2*x^2+..+C_n^n*x^n)+ +((C_n^01*x+C_n^1*x^2+C_n^2*x^3+..+C_n^n*x^{n+1})) ](https://tex.z-dn.net/?f=%28C_n%5E01%2BC_n%5E1%2Ax%5E1%2BC_n%5E2%2Ax%5E2%2B..%2BC_n%5En%2Ax%5En%29%2B+%0A%0A%2B%28%28C_n%5E01%2Ax%2BC_n%5E1%2Ax%5E2%2BC_n%5E2%2Ax%5E3%2B..%2BC_n%5En%2Ax%5E%7Bn%2B1%7D%29%29%0A)
соберем подобные слагаемые:
![C_n^01+x(C_n^1+C_n^0)+x^2(C_n^1+C_n^2)+...x^n(C_n^{n+1}+C_n^n)+x^{n+1}(C_n^n)](https://tex.z-dn.net/?f=C_n%5E01%2Bx%28C_n%5E1%2BC_n%5E0%29%2Bx%5E2%28C_n%5E1%2BC_n%5E2%29%2B...x%5En%28C_n%5E%7Bn%2B1%7D%2BC_n%5En%29%2Bx%5E%7Bn%2B1%7D%28C_n%5En%29)
теперь правило
![C_n^n+C_n^{n-1}=C_{n+1}^n; C_{n}^n=C_{n+1}^{n+1}](https://tex.z-dn.net/?f=C_n%5En%2BC_n%5E%7Bn-1%7D%3DC_%7Bn%2B1%7D%5En%3B+C_%7Bn%7D%5En%3DC_%7Bn%2B1%7D%5E%7Bn%2B1%7D)
преобразуем нашу сумму:
![C_n^01+x(C_{n+1}^1)+x^2(C_{n+1}^2)+...x^n(C_{n+1}^{n})+x^{n+1}(C_{n+1}^{n+1})=](https://tex.z-dn.net/?f=C_n%5E01%2Bx%28C_%7Bn%2B1%7D%5E1%29%2Bx%5E2%28C_%7Bn%2B1%7D%5E2%29%2B...x%5En%28C_%7Bn%2B1%7D%5E%7Bn%7D%29%2Bx%5E%7Bn%2B1%7D%28C_%7Bn%2B1%7D%5E%7Bn%2B1%7D%29%3D)
![= \Sigma_{k=0}^{n+1}(C_{n+1}^k*x^k)](https://tex.z-dn.net/?f=%3D+%5CSigma_%7Bk%3D0%7D%5E%7Bn%2B1%7D%28C_%7Bn%2B1%7D%5Ek%2Ax%5Ek%29)
Что и требовалось доказать
Дополнительно докажу:
![C_n^p+C_n^{p+1}=C_{n+1}^{p+1} ](https://tex.z-dn.net/?f=C_n%5Ep%2BC_n%5E%7Bp%2B1%7D%3DC_%7Bn%2B1%7D%5E%7Bp%2B1%7D%0A%0A)
-7 и 7
-4 и 4
-1 и 1
2 и -2
5 и -5
11 и -11
14 и -14
Решение представлено на фото
6x+4x+20-10x-20=0
0*x=0
Значить рівняння має безліч розв'язків