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`1^2 + 3^2 + 5^2 + (2n - 1)^2 = (n(4n^2 - 1))/3`  prove this using mathematical induction

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hiani | Student, Grade 11 | (Level 2) eNoter

Posted April 25, 2013 at 7:29 PM via web

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`1^2 + 3^2 + 5^2 + (2n - 1)^2 = (n(4n^2 - 1))/3`  prove this using mathematical induction

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tiburtius | High School Teacher | (Level 3) Associate Educator

Posted April 25, 2013 at 8:03 PM (Answer #1)

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First check for `n=1`

`1^2=(1cdot(4cdot1^2-1))/3`

`1=1`

So for `n=1` the identity holds. Now assume that the identity holds for all natural numbers greater or equal to `n` i.e.

`1^2+3^2+5^2+cdots+(2k-1)^2=(k(4k^2-1))/3,` `k leq n`             (1)

Let's now check for `k=n+1`

`1^2+3^2+5^2+cdots+(2n-1)^2+(2(n+1)-1)^2=`

Now by using assumption (1) we get

`(n(4n^2-1))/3+(2n+1)^2=`

`(4n^3-n)/3+4n^2+4n+1=`

`(4n^3-n+12n^2+12n+3)/3=`

`((n+1)(4(n+1)^2-1))/3`

which we were supposed to get. Hence our proof is done.

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