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December 21, 2014

December 21, 2014

Posted by **Jesse** on Thursday, June 10, 2010 at 12:33am.

(1)(2) + (2)(3) + ... n(n+1) =

n(n+1)(n+2)/3

complete the inductive step of its proof:

1(2) + 2(3) + ... k(k+1) + (k+1)(k+2)

I have no idea what to do.

- pre-calculus -
**Reiny**, Thursday, June 10, 2010 at 7:05amstep1

test for n=1

LS = (1)(2) = 2

RS = (1)(2)(3)/3 = 2, check!

step2

assume it is true for n=k, that is ...

(1)(2) + (2)(3) + ... k(k+1) =

k(k+1)(k+2)/3

step3

show that it must be true for n= k+1

that is ...

show

(1)(2) + (2)(3) + ... k(k+1) + (k+1)(k+2) =

(k+1)(k+2)(k+3)/3

LS = (1)(2) + (2)(3) + ... k(k+1) + (k+1)(k+2)

= [ (1)(2) + (2)(3) + ... k(k+1) ] + (k+1)(k+2)

= k(k+1)(k+2)/3 + (k+1)(k+2)

= k(k+1)(k+2)/3 + 3(k+1)(k+2)/3

= [ k(k+1)(k+2) + 3(k+1)(k+2) ]/3

= (k+1)(k+2) [k + 3]/3 , took out common factor

= (k+1)(k+2)(k+3)/3

= RS

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