Mathematics: I need help, can someone please help me

I need help, can someone please help me

Ответ:

miguelpoblano5673

09.02.2022

Mathematics

slope (3,1) using rise over run y-intercept -1

Step-by-step explanation:

Starting at your y-intercept -1 and going up three and 1 to the right you get to your next point on the graph

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Ответ:

francisco42002

18.11.2022

Mathematics

$2^2+4^2+6^2+...(2n)^2=\frac{2n(n+1)(2n+1)}{3};\ n\geq1\\\\chek\ for\ n=1:\\L=2^2=4;\ R=\frac{2\cdot1(1+1)(2\cdot1+1)}{3}=\frac{2\cdot2\cdot3}{3}=4\\L=R\\-----------------------\\
assumption\ for\ n=k\\2^2+4^2+6^2+...+(2k)^2=\frac{2k(k+1)(2k+1)}{3}\\-----------------------\\thesis\ for\ n=k+1\\2^2+4^2+6^2+...+(2k)^2+[2(k+1)]^2=\frac{2(k+1)(k+1+1)[2(k+1)+1]}{3}\\-----------------------$
$proff:\\L=2^2+4^2+6^2+...+(2k)^2+(2k+2)^2=\frac{2k(k+1)(2k+1)}{3}+(2k+2)^2\\\\=\frac{(2k^2+2k)(2k+1)}{3}+\frac{3(2k+2)^2}{3}=\frac{4k^3+2k^2+4k^2+2k+3(4k^2+8k+4)}{3}\\\\=\frac{4k^3+6k^2+2k+12k^2+24k+12}{3}=\boxed{\frac{4k^3+18k^2+26k+12}{3}}\\\\R=\frac{2(k+1)(k+1+1)[2(k+1)+1]}{3}=\frac{(2k+2)(k+2)(2k+2+1)}{3}\\\\=\frac{(2k^2+4k+2k+4)(2k+3)}{3}=\frac{(2k^2+6k+4)(2k+3)}{3}=\frac{4k^3+6k^2+12k^2+18k+8k+12}{3}\\\\=\boxed{\frac{4k^3+18k^2+26k+12}{3}}\\\\L=R$

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