Question

Using the method of mathematical induction to prove that equalities are true for values ​​of n indicated:

2²+4²+6²+...+(2n)²=
`(2n(n+1)(2n+1))/(3) ,n \geq 1.`

Answer 1

`2^2+4^2+6^2+...(2n)^2=(2n(n+1)(2n+1))/(3);\ n\geq1\\\\chek\ for\ n=1:\\L=2^2=4;\ R=(2\cdot1(1+1)(2\cdot1+1))/(3)=(2\cdot2\cdot3)/(3)=4\\L=R\\-----------------------\\ assumption\ for\ n=k\\2^2+4^2+6^2+...+(2k)^2=(2k(k+1)(2k+1))/(3)\\-----------------------\\thesis\ for\ n=k+1\\2^2+4^2+6^2+...+(2k)^2+[2(k+1)]^2=(2(k+1)(k+1+1)[2(k+1)+1])/(3)\\-----------------------`

`proff:\\L=2^2+4^2+6^2+...+(2k)^2+(2k+2)^2=(2k(k+1)(2k+1))/(3)+(2k+2)^2\\\\=((2k^2+2k)(2k+1))/(3)+(3(2k+2)^2)/(3)=(4k^3+2k^2+4k^2+2k+3(4k^2+8k+4))/(3)\\\\=(4k^3+6k^2+2k+12k^2+24k+12)/(3)=\boxed{(4k^3+18k^2+26k+12)/(3)}\\\\R=(2(k+1)(k+1+1)[2(k+1)+1])/(3)=((2k+2)(k+2)(2k+2+1))/(3)\\\\=((2k^2+4k+2k+4)(2k+3))/(3)=((2k^2+6k+4)(2k+3))/(3)=(4k^3+6k^2+12k^2+18k+8k+12)/(3)\\\\=\boxed{(4k^3+18k^2+26k+12)/(3)}\\\\L=R`