1) Базис индукции: n=1
![(2\cdot 1-1)^2=\dfrac{1\cdot(2\cdot1-1)(2\cdot 1+1)}{3}\\\\1=1](https://tex.z-dn.net/?f=%282%5Ccdot%201-1%29%5E2%3D%5Cdfrac%7B1%5Ccdot%282%5Ccdot1-1%29%282%5Ccdot%201%2B1%29%7D%7B3%7D%5C%5C%5C%5C1%3D1)
2) Предположим что и при n=k равенство верно
![1^2+2^2+3^2+...+(2n-1)^2=\dfrac{k(2k-1)(2k+1)}{3}](https://tex.z-dn.net/?f=1%5E2%2B2%5E2%2B3%5E2%2B...%2B%282n-1%29%5E2%3D%5Cdfrac%7Bk%282k-1%29%282k%2B1%29%7D%7B3%7D)
3) Индукционный переход: n = k+1
![\underbrace{1^2+2^2+3^2+...+(2k-1)^2}_{\dfrac{k(2k-1)(2k+1)}{3}}+(2(k+1)-1)^2=\dfrac{(k+1)(2k+1)(2k+3)}{3}](https://tex.z-dn.net/?f=%5Cunderbrace%7B1%5E2%2B2%5E2%2B3%5E2%2B...%2B%282k-1%29%5E2%7D_%7B%5Cdfrac%7Bk%282k-1%29%282k%2B1%29%7D%7B3%7D%7D%2B%282%28k%2B1%29-1%29%5E2%3D%5Cdfrac%7B%28k%2B1%29%282k%2B1%29%282k%2B3%29%7D%7B3%7D)
Докажем теперь равенство, а именно покажем что левая часть равна правой части.
![\dfrac{k(2k-1)(2k+1)}{3}+(2(k+1)-1)^2=\dfrac{k(2k-1)(2k+1)}{3}+(2k+1)^2=\\ \\ =\dfrac{k(2k-1)(2k+1)+3(2k+1)^2}{3}=\dfrac{(2k+1)(k(2k-1)+3(2k+1))}{3}=\\ \\ =\dfrac{(2k+1)(2k^2-k+6k+3)}{3}=\dfrac{(2k+1)(2k^2+2k+3k+3)}{3}=\\ \\ =\dfrac{(2k+1)((2k(k+1)+3(k+1))}{3}=\dfrac{(2k+1)(k+1)(2k+3)}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk%282k-1%29%282k%2B1%29%7D%7B3%7D%2B%282%28k%2B1%29-1%29%5E2%3D%5Cdfrac%7Bk%282k-1%29%282k%2B1%29%7D%7B3%7D%2B%282k%2B1%29%5E2%3D%5C%5C%20%5C%5C%20%3D%5Cdfrac%7Bk%282k-1%29%282k%2B1%29%2B3%282k%2B1%29%5E2%7D%7B3%7D%3D%5Cdfrac%7B%282k%2B1%29%28k%282k-1%29%2B3%282k%2B1%29%29%7D%7B3%7D%3D%5C%5C%20%5C%5C%20%3D%5Cdfrac%7B%282k%2B1%29%282k%5E2-k%2B6k%2B3%29%7D%7B3%7D%3D%5Cdfrac%7B%282k%2B1%29%282k%5E2%2B2k%2B3k%2B3%29%7D%7B3%7D%3D%5C%5C%20%5C%5C%20%3D%5Cdfrac%7B%282k%2B1%29%28%282k%28k%2B1%29%2B3%28k%2B1%29%29%7D%7B3%7D%3D%5Cdfrac%7B%282k%2B1%29%28k%2B1%29%282k%2B3%29%7D%7B3%7D)
Что и требовалось доказать
1) х-4х+11+9-2х=-5х+20
2) 10-9а+6+5а-16=-4а
У=-10х-9
у=-24х+19
-10х-9=-24х+19
14х=28
х=2
у=-10*2-9=-24*2+19=-29
(.) (2;-29)
Очень просто!
Расстояние S = скорость V * время t
Отсюда t = S / V = 120 / u
Да, это функция, аргумент - скорость u.