We will use the Principle of Mathematical Induction to show that 12+22+32+42+...+(2n)2=3n(2n+1)(4n+1)
1) Show that it is true for n=1: 12+22=5=31×3×5
2) Assume it is true for n=k: 12+22+...+(2k)2=3k(2k+1)(4k+1)
Now, prove it is true for k+1:
12+22+...+(2k+2)2=(12+...+(2k)2)+(2k+1)2+(2k+2)2=3k(2k+1)(4k+1)+(2k+1)2+(2k+2)2=3k(2k+1)(4k+1)+3(2k+1)2+12(k+1)2=38k3+30k2+37k+15=3(k+1)(2(k+1)+1)(4(k+1)+1)
It is true.
So, 12+22+32+42+...+(2n)2=3n(2n+1)(4n+1).
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