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Question

Verify that for all n1 , 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


A

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Solution

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For any integer n1 , 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 k1 , 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 n1 ,Pn holds.


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