LHS = (p+1)^2 + (p+1) = p^2 +2p +1RHS= 2^(p+1) = 2^p *2 = 2(p^2 + p)

[Firben]

Prove by induction that n^2 + n ≤ 2^n

for all integers n≥5

How i did:

Case(1)

Suppose that n = 5

LHS = 5^2 +5 = 30

RHS = 2^5 = 32

30 ≤ 32

Ok LHS ≤ RHS

Case (2)

Suppose that's true for n=p≥5. Show that its true for n = p+1

What should i do next ? I had a memory loss here :(

 

[Fredrik]

Just write down (p+1)^2+(p+1) and work with that expression until you see that it's \leq 2^{p+1}. You will have to use the assumption p^2+p\leq 2^p.

 

[Firben]

LHS(p+1) = (p+1)^2 + p+1 = p^2 + 2p + 1 + p + 1 = p^2 +3p + 2

How can i continue ?

 

[Mentallic]

LHS(p+1) = (p+1)^2 + p+1 = p^2 + 2p + 1 + p + 1 = p^2 +3p + 2

How can i continue ?

Well, what's your assumption?

 

[Firben]

Yes, i think so

 

[Fredrik]

No, he asked "what's your assumption?", and you were supposed to answer "that the formula I want to prove for all n≥5 holds when n=p".