Verify that for all n≥1 , the sum of the squares of the first 2n positive integers is given by the formula 12+22+32+.....(2n)2=n(2n+1)(4n+1)3
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For any integer n≥1 , let pn be the statement that
12+22+32+.....(2n)2=n(2n+1)(4n+1)3.
Base case––––––––––– : The statement P1 says that
12+22=(1)(2(1)+1)(4(1)+1)3=3(5)3=5,
This is true.
Inductive step––––––––––––––––: Fix k≥1 , and suppose that Pk holds, that is,
12+22+32+.....+(2k)2=k(2k+1)(4k+1)3.
It remains to show that Pk+1 holds, that is,
12+22+32+.....+(2(k+1))2=(k+1)(2(k+1)+1)(4(k+1)+1)3.
12+22+32+.....+(2(k+1))2=12+22+32+.....(2k+2)2
=12+22+32+.....+(2k)2+(2k+1)2+(2k+2)2
=k(2k+1)(4k+1)3+(2k+1)2+(2k+2)2 (by Pk)
=k(2k+1)(4k+1)3+3(2k+1)2+3(2k+2)23
=k(2k+1)(4k+1)+3(2k+1)2+3(2k+2)23
=k(8k2+6k+1)+3(4k2+4k+1)+3(4k2+8k+4)3
=(8k3+6k2+k)+(12k2+12k+3)+(12k2+24k+12)3
=8k3+30k2+37k+153
on the otherside of Pk+1,
=(k+1)(2(k+1)+1)(4(k+1)+1)3=(k+1)(2k+2+1)(4k+4+1)3
= (k+1)(2k+3)(4k+5)3
= (2k2+5k+3)(4k+5)3
=8k3+30k2+37k+153.
Therefore Pk+1 holds
Thus, by the principle of mathematical induction, for all n≥1 ,Pn holds.