precalculus
posted by Jesse on .
If
(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.

step1
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