Use the Induction to Prove that.

Question

I need ur help with this question.

The question states that: Use Induction TO Prove that

• 1*2+2*3+...+n(n+1) = n(n+1)(n+2)/3

So, so far I used to the basis step to show that this statement is indeed true.

• For n = 1, we have 1(1+1) = 1(1+1)(1+2)/3
• 2 = 2 which is true

Now the Inductive Step:

• Assume that P(n): 1*2+2*3+...+n(n+1) = n(n+1)(n+2)/3

Now from then I have no idea what to do. So can someone help me finish this unfinished question.

Thank You.

Answer 1

OK, next step: Proving for k+1. Let's take $$1\cdot2+2\cdot3+....k(k+1)+(k+1)(k+2)=\frac{(k+1)(k+2)(k+3)}{3}$$

Using the induction step we have... $$\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$$

Simplifying we get $$\frac{k^3+3k^2+2k}{3}+\frac{3k^2+9k+6}{3}$$ Which equals $$\frac{k^3+6k^2+11k+6}{3}$$

Factor and we have $$\frac{(k+1)(k+2)(k+3)}{3}$$

Answer 2

First, show that this is true for $n=1$:

$\sum\limits_{k=1}^{1}k(k+1)=1(1+1)(1+2)/3$

Second, assume that this is true for $n$:

$\sum\limits_{k=1}^{n}k(k+1)=n(n+1)(n+2)/3$

Third, prove that this is true for $n+1$:

$\sum\limits_{k=1}^{n+1}k(k+1)=$

$\color\red{\sum\limits_{k=1}^{n}k(k+1)}+(n+1)(n+2)=$

$\color\red{n(n+1)(n+2)/3}+(n+1)(n+2)=$

$(n+1)(n+2)(n+3)/3$

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Please note that the assumption is used only in the part marked red.