Show by Induction, but stuck on Induction Step (Don't know how to implement N = K+1)

Question

Prove by Induction that $$ 1\cdot 2\cdot3 + 2\cdot3\cdot4 + \cdots + n \cdot(n+1)\cdot(n+2) = \frac{n (n+1)(n+2) (n+3)}{4} $$

For my basis step, I set $$n=0$$ This leaves me with zero on both LHS and RHS, so I proceed. Induction Hypothesis consists of plugging in $N = K$ assuming the basis step applies for $$ k \leq n $$

After this, I am stuck on the Induction Step. I try to solve for $n = k+1$, but I don't know what to do.

Due to requests asking me to show what I have tried: $$ (n+1)(n+2)(n+3) = \frac{(n+1)(n+2)(n+3)(n+4)}{4} $$ I don't know where to go from here. I don't think it is correct anyways.

Answer 1

By assumption of the induction we obtain: $$1\cdot2\cdot3+\dots+n(n+1)(n+2)+(n+1)(n+2)(n+3)=$$ $$=\frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3)=$$ $$=(n+1)(n+2)(n+3)\left(\frac{n}{4}+1\right)=\frac{(n+1)(n+2)(n+3)(n+4)}{4}.$$ Can you end it now?

Answer 2

Just sum both sides $(n+1)(n+2)(n+3)$ yo your Induction Hypothesis. So you get:

$(1)(2)(3) + ... + n(n+1)(n+2) + (n+1)(n+2)(n+3) = \frac{(n)(n+1)(n+2)}{4}+(n+1)(n+2)(n+3)=\frac{(n)(n+1)(n+2)+4(n+1)(n+2)(n+3)}{4}=\frac{(n+1)(n+2)(n+3)(n+4)}{4}$

Answer 3

@user758853.

Your question is actually similar to this:

Prove that $\sum_{i=1}^{n} i \times (i+1) = \frac{n(n+1)(n+2)}{3}$

To solve this, you multiply both sides by 3, then do the following

$ 1 \times 2 \times 3 = 1 \times 2 \times (3-0) = 1 \times 2 \times 3 - 0$

$ 2 \times 3 \times 3= 2 \times 3 \times (4-1) = 2 \times 3 \times 4 - 1\times 2 \times 3 $

$ 3 \times 4 \times 3= 3 \times 4 \times (5-2) = 3 \times 4 \times 5 -2 \times 3 \times 4 $

...

Do you recognize the pattern now?

By using the same method you do not have to use induction anymore, just multiply each sides of your equation by 4 instead of 3 and you will find the given equation correct.

EDITS

I think mathematical induction also applies, and if you don't know how to do it, just follow these steps:

• Apply your induction hypothesis to the first $n$ factors of the sum
• Add the $(n+1)th$ factor to the LHS, and write what you need to prove on the RHS.
• Try reducing both sides to the same denominator and expand both sides then you should see it apparently true.

Q.E.D