Expert: Steve Holleran - 5/19/2007

Question
Hi i am really stuck on these questions. Can you show me how they are done?
1. Use induction to prove that for any real number a>=0:(1+a)^n>=(1+na) for all n>=1.
2. Prove by induction that:
1x2+2x3+...+n(n+1)=1/3n(n+1)(n+2)
for all integers n>0.


Answer
Hi Scott,

Well, maybe the reason you are really stuck is because I can't see how the first of these can be true.

Okay, just to be sure we're on the same wavelength, in using induction, you want to do these two steps:

I.  Show the proposition is true for n = 1.

II. Show that assuming the proposition true for n = k, then the proposition is true for n = k + 1.

Alright then, for #1,

P(1) :   (1 + a)^1 = (1 + 1*a)----> 1 + a = 1 + a    TRUE

P(k)-->P(k+1) : Assuming (1 + a)^k = (1 + k * a),

show (1 + a)^(k+1) = (1 + (k+1)*a).

My problem is that this is not even true for n = 2:

    (1 + a)^2 does not = (1 + 2*a)   because

    (1 + a)^2 = 1 + 2a + a^2, and this is true ONLY when a=0.  So, something is seriously wrong here.

On #2,  I had a lot of trouble with your notation.  First, I thought you meant

 1*2 + ... + n(n+1) = 1 / [3n * (n+1) * (n+2)

But I saw that couldn't work.  Then, I thought you meant

 1*2 + ... + n(n+1) = (1/3n) * (n+1) * (n+2),  but that didn't work either.  I hope the correct notation is:

 1*2 + ... + n(n+1) = (1/3) * n * (n+1) * (n+2)

If so, then here we go:

P(1):   1*2 = (1/3) * (1) * (2) * (3)

        2  = (1/3) * 6 = 2  TRUE

P(k)--> P(k+1):

Assuming, 1*2 +... + k(k+1) = (1/3) * k * (k+1) * (k+2)

Show 1*2 + ... + k(k+1) + (k+1)(k+2)
    
           = (1/3) * (k+1) * (k+2) * (k+3).

Okay, in the equation immediately above, on the left side
all the terms from 1*2  to k(k+1) can be replaced with the assumption. Then on the left you have:

(1/3) * k * (k+1) * (k+2) + (k+1)(k+2)

Now, you have a common factor of (k+1)(k+2), so this now becomes:

(k+1)(k+2) *[(1/3) * k)+ 1]

= (k+1)(k+2) * [ (k/3) + 1]

= (k+1)(k+2) * [ (k+3) / 3]  (common denom in the bracket]

= (1/3) * (k+1)(k+2)(k+3)

which is what you want to prove.

Hope this helps out.  Let me know if I misinterpreted that first one.

Steve Holleran