Question

12 pts) Use Mathematical Induction to prove that a_n=n³+5n is divisible by 6 when ever n≥0. You may explicitly use without proof the fact that the product n(n+1) of consecutive integers n and n+1 is always even, that is, you must state where you use this fact in your proof.Write in complete sentences since this is an induction proof and not just a calculation. Hint:Look up Pascal’s triangle.

(a)

Verify the initial case n= 0.

(b)

State the induction hypothesis.

(c)

Perform the induction

Answer 1

We need to prove that n³ +5n is divisible by 6 when n≥0.

(a) Initial case : n=0

When n=0,

n³ + 5n = (0)³ +(5*0) =0

0 is divisible by 6. Hence initial case is satisfied

(b) Inductive step: Assume that k³ + 5k is divisible by 6 for some k N.........(1)

(c) We need to prove (k+1)³ + 5(k+1) is divisible by 6

(k+1)³ +5(k+1)= k³ +3k² +3k+1 +5k +5 =k³ + 3k² +8k +6

Rearranging the terms,

(k+1)³ +5(k+1)=k³ +5k +3k+3k² +6

From (1), we can state that k³ +5k=6m for some integer m.

Hence,

(k+1)³ +5(k+1)=6m +3k+3k² +6

(k+1)³ +5(k+1)= 6(m+1)+ 3k+3k² ..........(2)

Let k be an even number. Hence k=2i for some integer i

(2) becomes,

(k+1)³ +5(k+1)=6(m+1)+ 3k+3k² =  6(m+1)+ 3*2i +3*(2i)²

(k+1)³ +5(k+1)=  6(m+1)+6i+ 12i²   

  (k+1)³ +5(k+1)= 6(m+1+i+2i²), which is divisible by 6

Let k be an even number. Hence k=2i+1 for some integer i

  (2) becomes,

(k+1)³ +5(k+1)=6(m+1)+ 3k+3k² = 6(m+1) +3(2i+1)+3(2i+1)²

  (k+1)³ +5(k+1)=6(m+1)+6i+3+3(4i²+4i+1)

  (k+1)³ +5(k+1)=6(m+1)+6i+3+12i² +12i +3

(k+1)³ +5(k+1)=6(m+1)+6(2i² +3i+1)

(k+1)³ +5(k+1)=6(2i² + 3i + 2 + m ), which is divisible by 6

Hence, by induction we prove that   n³ +5n is divisible by 6 when n≥0.  

The fact that the product n(n+1) of consecutive integers n and n+1 is always even is not used in this proof.