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September 20, 2014

September 20, 2014

Posted by **Josh** on Friday, July 29, 2011 at 4:29pm.

2 + 6 + 10 + . . . + (4n – 2) = 2n^2

is true

- Math -
**Damon**, Friday, July 29, 2011 at 5:38pmn an sum n

1 2 2

2 6 8

3 10 18

4 14 32

n (4n-2) 2n^2

(n+1) (4(n+1)-2) [2n^2+ (4(n+1)-2)] or hopefully 2(n+1)^2

[2n^2+ (4(n+1)-2)] = 2n^2+4n +2

and

2(n+1)^2 = 2(n^2+2n+1) = 2n^2+4n+2 done

2n^2+ (4(n+1)-2)

- Math -
**Reiny**, Friday, July 29, 2011 at 5:46pmUsing the 3 step process:

1. test for n=1

LS = 2

RS = 2(1^2) = 2

2. assume it true for n=k

that is ....

2+4+6 + ... + (4k-2) = 2k^2

3 . prove it is then true for =k+1

or

2+4+6+ ... + (4k-2) + 4(k+1)-2 = 2(k+1)2

LS = [2+4+6+...+ 4k-2 ] + 4(k+1)-2

= 2k^2 + 4k + 4 - 2

= 2(k^2 + 2k + 1)

= 2(k+1)^2

= RS

QED

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