В равенстве слева сумма имеет общий член ![a_n=n(2n+1)](https://tex.z-dn.net/?f=a_n%3Dn%282n%2B1%29)
1) Базис индукции: ![n=1](https://tex.z-dn.net/?f=n%3D1)
![1\cdot (2\cdot 1+1)=\dfrac{1\cdot (1+1)\cdot (4\cdot 1+5)}{6}~~~\Rightarrow~~~ 3=3](https://tex.z-dn.net/?f=1%5Ccdot%20%282%5Ccdot%201%2B1%29%3D%5Cdfrac%7B1%5Ccdot%20%281%2B1%29%5Ccdot%20%284%5Ccdot%201%2B5%29%7D%7B6%7D~~~%5CRightarrow~~~%203%3D3)
2) Предположим, что и для
верно равенство
![1\cdot 3+2\cdot 5+...+k(2k+1)=\dfrac{k(k+1)(4k+5)}{6}](https://tex.z-dn.net/?f=1%5Ccdot%203%2B2%5Ccdot%205%2B...%2Bk%282k%2B1%29%3D%5Cdfrac%7Bk%28k%2B1%29%284k%2B5%29%7D%7B6%7D)
3) Индукционный переход: ![n=k+1](https://tex.z-dn.net/?f=n%3Dk%2B1)
![\underbrace{1\cdot 3+2\cdot 5+...+k(2k+1)}_{\frac{k(k+1)(4k+5)}{6}}+(k+1)(2k+3)=\dfrac{(k+1)(k+2)(4k+9)}{6}\\ \\\\\dfrac{k(k+1)(4k+5)}{6}+(k+1)(2k+3)=\dfrac{(k+1)(k+2)(4k+9)}{6}\\ \\ \\ (k+1)\left(\dfrac{k(4k+5)}{6}+2k+3\right)=\dfrac{(k+1)(k+2)(4k+9)}{6}\\ \\ \\ (k+1)\cdot \dfrac{4k^2+5k+12k+18}{6}=\dfrac{(k+1)(k+2)(4k+9)}{6}\\ \\ \\ \dfrac{(k+1)(4k^2+17k+18)}{6}=\dfrac{(k+1)(k+2)(4k+9)}{6}\\ \\ \\ \dfrac{(k+1)(k+2)(4k+9)}{6}=\dfrac{(k+1)(k+2)(4k+9)}{6}](https://tex.z-dn.net/?f=%5Cunderbrace%7B1%5Ccdot%203%2B2%5Ccdot%205%2B...%2Bk%282k%2B1%29%7D_%7B%5Cfrac%7Bk%28k%2B1%29%284k%2B5%29%7D%7B6%7D%7D%2B%28k%2B1%29%282k%2B3%29%3D%5Cdfrac%7B%28k%2B1%29%28k%2B2%29%284k%2B9%29%7D%7B6%7D%5C%5C%20%5C%5C%5C%5C%5Cdfrac%7Bk%28k%2B1%29%284k%2B5%29%7D%7B6%7D%2B%28k%2B1%29%282k%2B3%29%3D%5Cdfrac%7B%28k%2B1%29%28k%2B2%29%284k%2B9%29%7D%7B6%7D%5C%5C%20%5C%5C%20%5C%5C%20%28k%2B1%29%5Cleft%28%5Cdfrac%7Bk%284k%2B5%29%7D%7B6%7D%2B2k%2B3%5Cright%29%3D%5Cdfrac%7B%28k%2B1%29%28k%2B2%29%284k%2B9%29%7D%7B6%7D%5C%5C%20%5C%5C%20%5C%5C%20%28k%2B1%29%5Ccdot%20%5Cdfrac%7B4k%5E2%2B5k%2B12k%2B18%7D%7B6%7D%3D%5Cdfrac%7B%28k%2B1%29%28k%2B2%29%284k%2B9%29%7D%7B6%7D%5C%5C%20%5C%5C%20%5C%5C%20%5Cdfrac%7B%28k%2B1%29%284k%5E2%2B17k%2B18%29%7D%7B6%7D%3D%5Cdfrac%7B%28k%2B1%29%28k%2B2%29%284k%2B9%29%7D%7B6%7D%5C%5C%20%5C%5C%20%5C%5C%20%5Cdfrac%7B%28k%2B1%29%28k%2B2%29%284k%2B9%29%7D%7B6%7D%3D%5Cdfrac%7B%28k%2B1%29%28k%2B2%29%284k%2B9%29%7D%7B6%7D)
Равенство выполняется для всех натуральных n. Что и требовалось доказать.
5 sin 3x -5=0
5 sin 3x=5
sin 3x=5/5
sin 3x =0
3x=pi k, k∈Z
x=pi/3 k, k∈Z
Ответ: pi/3 k, k∈Z
(n²+n)/2=55
n²+2=55*2
n²+n=110
n²+n-110=0
D=441=21²
n₁=(-1+21)/2=10
n₂=(-1-21)/2=-11∉N
n=10
<span>tg (-t)×cos t - sin (4П-t)=-tg (t)*cos t-(-sint)=tg (t)*cos(t)+sin(t)=sin t/cost*cost+sint=sint+sint=2sint </span>