1² + 3² + 4² + 4(n - 1)² = ¹/₃n(2n - 1)(2n + 1)
1² + 3² + 4² + (2n - 2)² = ¹/₃n(2n - 1)(2n + 1)
1 + 9 + 16 + (2n - 2)(2n - 2) = ¹/₃n(2n(2n + 1) - 1(2n + 1))
10 + 16 + (2n(2n - 2) - 2(2n - 2)) = ¹/₃n(2n(2n) + 2n(1) - 1(2n) - 1(1) 16 + (2n(2n) - 2n(2) - 2(2n) + 2(2)) = ¹/₃n(4n² + 2n - 2n - 1)
26 + (4n² - 4n - 4n + 4) = ¹/₃n(4n² - 1)
26 + (4n² - 8n + 4) = ¹/₃n(4n² - 1)
26 + 4n² - 8n + 4 = ¹/₃n(4n²) - ¹/₃n(1)
4n² - 8n + 4 + 26 = 1¹/₃n³ - ¹/₃n
4n² - 8n + 30 = 1¹/₃n³ - ¹/₃n
+ ¹/₃n + ¹/₃n
4n² - 7²/₃n + 30 = 1¹/₃n³
-1¹/₃n³ + 4n² - 7²/₃n + 30 = 0
-3(-1¹/₃n³ + 4n² - 7²/₃n + 30) = -3(0)
-3(-1¹/₃n³) - 3(4n²) - 3(-7²/₃n) - 3(30) = 0
4n³ - 12n² + 23n - 90 = 0
Answer:
1) 8
2) 1
Step-by-step explanation:
Answer:
Step-by-step explanation:
Use the formula,
distance = √[(x1-x2)^2 + (y1-y2)^2]
Answer:
Step-by-step explanation:
![2(x-2)^2=8(7+y)\\2(x-2)^2=56+8y \Rightarrow y={1\over{8}}[2(x-2)^2-56]={1\over{4}}(x-2)^2-7\\finally\\y={1\over{4}}(x-2)^2-7\\let ~change~x~and~y\\x={1\over{4}}(y-2)^2-7\\x+7={1\over{4}}(y-2)^2\\4(x+7)=(y-2)^2\\y-2=\pm\sqrt{4(x+7)}\\\\y=2\pm\sqrt{4x+28}](https://tex.z-dn.net/?f=2%28x-2%29%5E2%3D8%287%2By%29%5C%5C2%28x-2%29%5E2%3D56%2B8y%20%5CRightarrow%20y%3D%7B1%5Cover%7B8%7D%7D%5B2%28x-2%29%5E2-56%5D%3D%7B1%5Cover%7B4%7D%7D%28x-2%29%5E2-7%5C%5Cfinally%5C%5Cy%3D%7B1%5Cover%7B4%7D%7D%28x-2%29%5E2-7%5C%5Clet%20~change~x~and~y%5C%5Cx%3D%7B1%5Cover%7B4%7D%7D%28y-2%29%5E2-7%5C%5Cx%2B7%3D%7B1%5Cover%7B4%7D%7D%28y-2%29%5E2%5C%5C4%28x%2B7%29%3D%28y-2%29%5E2%5C%5Cy-2%3D%5Cpm%5Csqrt%7B4%28x%2B7%29%7D%5C%5C%5C%5Cy%3D2%5Cpm%5Csqrt%7B4x%2B28%7D)
