How to demonstrate that #n^3+2n# is divisible by #3# for #n = 1,2, cdots# ?

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Answer 1 · 2017-12-06T14:29:52

see below

we can do this by induction

#T_n=n^3+2n#

#(1)# Verify for #n=1#

#T_1=1^3+2xx1=3#

#:.T_1" "#is divisible by #3#

#(2#) assume #T_k=3p, p in NN#

#:.n^3+2n=3p#

#(3) " to prove " T_k=>T_(k+1)#

#T_(k+1)=(k+1)^3+2(k+1)#

#T_(k+1)=k^3+3k^2+3k+1+2k+2#

#T_(k+1)=k^3+2k +3k^2+3k+3#

#because n^3+2n=3p# by assumption

#=>T_k=3p+3(k^2+k+1)#

#T_k=3(p+k^2+k+1)#

#:. T_k" "#is a multiple of #3#

#:.T_k=>T_(k+1)#

#(3)#conclusion

true for #n=1#

#T_k=>T_(k+1)#

#:. T_1=>T_2#

#T_2=>T_3#etc.

#:.n^3+2n" "#is divisible by#" "3, AAn in NN#

Answer 2 · 2017-12-06T16:32:33

See below.

#n(n+1)(n+2) = n^3+3n^2+2n#

We know that #n(n+1)(n+2)# is divisible by #3#

then

#n^3+3n^2+2n# is also divisible by #3#

but clearly #3n^2# is divisible by #3#

then #n^3+2n# also is divisible by #3#