# Proof by induction; inequality $1\cdot3+2\cdot4+3\cdot5+\dots+n(n+2) \ge \frac{n^3+5n}3$

Ok so I'm kind of struggling with this: The question is:

"Use mathematical induction to prove that 1*3 + 2*4 + 3*5 + ··· + n(n + 2) ≥ (1/3)(n^3 + 5n) for n≥1"

Okay, so P(1) is true as 1(1+2)=3 and (1/3)(1^3 + 5)=2 Assuming P(k) is true for k≥1 gives:

1*3 + 2*4 + 3*5 + ... + k(k+2) ≥ (1/3)(k^3 + 5k)

And we want to show that P(k+1) is true, that is;

1*3 + 2*4 + 3*5 + ... + (k+1)(k+3) ≥ (1/3)((k+1)^3 + 5(k+1)) (inequality 1)

This is where I'm not sure if I'm thinking along the right lines or not. Surely, by P(k) and the rules for inequalities we get:

(1*3 + 2*4 + 3*5 + ... + k(k+2)) + (k+1)(k+3) ≥ (1/3)(k^3 + 5k) + (k+1)(k+3) (inequality 2)

but the right hand side in this inequality does not equal (1/3)((k+1)^3 + 5(k+1)) which is what I am trying to show. There's obviously a flaw in my thinking somewhere so any help would be greatly appreciated. Thanks

Edit: As far as I understand it the right hand side of inequality 1 should equal the right hand side of inequality 2 but this isn't the case when expanded

• because they are not equal. you should show that one is greater than the other – benji Feb 13 '15 at 20:12
• If your know $x \geq y$ and you find $y \geq z$, then you've established $x \geq z$. You'll find the expansion of the right-hand side of $(2)$ is greater than the right-hand side of $(1)$. So you simply need to show that. – Namaste Feb 13 '15 at 20:26
• I don't understand why they wouldn't be equal? I thought that if x>y then (x+a)>(y+a) which is how inequality 2 comes about – Michael Feb 13 '15 at 20:37

## 2 Answers

Assume that statement is true for number unto k. We need to show that it is true for k+1.

$$1*3+...+k(k+2)+(k+1)(k+3)\geq \frac{k^3+5k}{3}+(k+1)(k+3)=\frac{k^3+3k^2+17k+9}{3}\geq\frac{k^3+3k^2+8k+6}{3}=\frac{(k+1)^3+5(k+1)}{3}$$

for last inequality expand both sides and see that

$$\frac{(k+1)^3+5(k+1)}{3}=\frac{k^3+3k^2+8k+6}{3}$$

and $$\frac{k^3+5k}{3}+(k+1)(k+3)=\frac{k^3+3k^2+17k+9}{3}$$

put it in original inequation and get the result

• I'm not sure how this proves the result? I know the equations are different but surely we're trying to show that the last inequality is the same as the RHS of P(k+1)? I'm confused lol – Michael Feb 13 '15 at 20:28
• Just forget the middle part and see far LHS $\geq$ far RHS. Do you see the statement P(k+1)? – Harish Feb 13 '15 at 20:32

Here's the meat of the argument: \begin{align} 1\cdot3+2\cdot 4+\cdots+(k+1)(k+3)&\geq \frac{k^3+5k}{3}+(k+1)(k+3)\tag{ind. hyp.}\\[1em] &= \frac{k^3+5k+3(k^2+4k+3)}{3}\tag{simplify}\\[1em] &= \frac{k^3+3k^2+17k+9}{3}\tag{simplify}\\[1em] &\geq \frac{k^3+3k^2+8k+6}{3}\tag{since $k\geq 1$}\\[1em] &= \frac{(k+1)^3+5(k+1)}{3}. \end{align} The last equality is what you want. The key is in realizing that since $k\geq 1$ you can obtain the desired expression on the right-hand side to show that $$1\cdot3+2\cdot 4+\cdots+(k+1)(k+3) \geq\frac{(k+1)^3+5(k+1)}{3}.$$

• ahhh I see now lol. Thanks for the explanation! – Michael Feb 13 '15 at 20:41
• @Michael You're welcome 8^) – Daniel W. Farlow Feb 13 '15 at 20:42