How to demonstrate this with mathematical induction?#P(n):n^3+5n# divide with 6 is true #forall ninNN#.

3 Answers
Daniel L.
Apr 24, 2017

See explanation.

Explanation:

To prove this identity using mathematical induction we have to follow these steps:

  1. Check the identity for #n=1#: #P(1)=1^3+5*1=6#. The result is a multiple of #6#, so the thesis is true.
  2. Assume that it is true for #n=k#, so: #EE{a in ZZ} k^3+5k=6a#
  3. Next step is to prove (using the assumption) that the thesis is true for #n=k+1#

Proof

#n=k+1#

#(k+1)^3+5(k+1)=k^3+3k^2+3k+1+5k+5=#
#=k^3+5k+3k^2+3k+6#

Now we can simplify the first 2 term using the assumption from point 2.

#k^3+5k+3k^2+3k+6=6a+3k^2+3k+6=#

#6a+3(k^2+k+2)=6a+3(k(k+1)+2)#

The first component of the sum is divisible by #6# (from the assumption). We have to show that the second expression is also divisible by #6#.

It is clearly divisible by #3#. As a sum of an even number and a product of 2 consecutive natural numbers the expression in brackets is an even number. Therfore the whole expression is divisible by #6#.

This concludes the proof.

sjc
Apr 24, 2017

see below

Explanation:

1) test for P(1)

#P(1):1^3+5xx1=1+5=6#

#:.P(1)=0("mod 6")#

#:." true for " n=1#

2) Assume true #n=k#

#P(k)=k^3+5k#

3) Prove for #k+1#

#(k+1)^3+5(K+1)=k^3+3k^2+3k+1+5k+5#

#P(k+1)=k^3+3k^2+8k+6#

#=>P(k+1)=(k^3+5k)+3(k^2+k++6)#

#(k^3+5k)=6p " by assumption"#

#=>P(k+1)=6p+3(k^2+k+6)#

#3(k^3+k++6)=0 ( mod 3)#

#(k^3+k+6) " for k "odd#

#k=2q-1=>(k^2+k+6)=(2q-1)^2+(2q-1)+6#

#=4q^2+2q^2+6=2(2q^2+2q^2+3)=0(mod2)#

so#" " 3(k^2+k+6)=6m#

#:.P(k+1) =6(p+m)=0(mod6) " for k "odd"#

#k " even"=k=2t#

#=>(k^2+k+6)=(2t)^2+2t+6=4t^2+2t+6#

#=2(2t^2+t+3)=2s#

#:.P(k)=6(p+s)=0(mod6) " for k even"#

#P(k)=>P(k+1) AAkinNN#

but true for #P(1)#

#:." by induction " P(1)=>P(2)=>P(3).... #

#" so true "AAninNN#

Cesareo R.
Apr 24, 2017

See a non inductive proof.

Explanation:

To compare, a non inductive proof

We know that

#n(n+1)(n+2)# is divisible by #6# because one of #n, n+1, n+2# is divisible by #3# and also at least one of them is even.

so #n(n+1)(n+2) equiv 0 mod 6#

but

#n^3+5n = n(n+1)(n+2) -3n(n-1)#

and here #3n(n-1) equiv 0 mod 6#

so

#n^3+5n equiv 0 mod 6#

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