# Thread: Need help on Mathematical Induction question

1. ## Need help on Mathematical Induction question

For statement

1*3 + 2*4 + 3-5 ... + (n-1)*(n+1) = (n(n-1)(2n+5)/6 )for all integers >= 2.

Base step(B)

P(2) is true since (2-1)x(2+1) = (2(2-1)(2(2)+5)/6)
3 = 3
thus P(2) is true.

Inductive step (R)

Assume P(n) is true for some n E N.
Prove that (n) x (n+2) = [(n+1)(n)(2n+1 +5)/6] is true. <-- is this correct? it looks wierd for some reason.

(n-1)x(n+1) + n x (n+2) = (n(n-1)(2n+5)/6 ) + n(n+2)
= ((n(n-1)(2n+5) + 6[n(n+2)]/6)

and this is where i got stuck. i durno how to simplify the equation could someone please tell me how to solve this problem?

2. Originally Posted by chogz

For statement

1*3 + 2*4 + 3-5 ... + (n-1)*(n+1) = (n(n-1)(2n+5)/6 )for all integers >= 2.

Base step(B)

P(2) is true since (2-1)x(2+1) = (2(2-1)(2(2)+5)/6)
3 = 3
thus P(2) is true.

Inductive step (R)

Assume P(n) is true for some n E N.
Prove that (n) x (n+2) = [(n+1)(n)(2n+1 +5)/6] is true. <-- is this correct? it looks wierd for some reason.

(n-1)x(n+1) + n x (n+2) = (n(n-1)(2n+5)/6 ) + n(n+2)
= ((n(n-1)(2n+5) + 6[n(n+2)]/6)

and this is where i got stuck. i durno how to simplify the equation could someone please tell me how to solve this problem?
What you need to do is show that

$\displaystyle (1\cdot3)+(2\cdot4)+(3\cdot5)+\dots+[(n-1)\cdot(n+1)]+[n\cdot(n+2)]=\frac{(n+1)(n)(2n+7)}{6}$

But,

$\displaystyle \underbrace{(1\cdot3)+(2\cdot4)+(3\cdot5)+\dots+[(n-1)\cdot(n+1)]}_{\frac{n(n-1)(2n+5)}{6}}+[n\cdot(n+2)]=\frac{(n+1)(n)(2n+7)}{6}$

So all you need to do now is get the left side to look like the right side:

$\displaystyle \frac{(n-1)(n)(2n+5)}{6}+n(n+2)=\frac{n(n+1)(2n+7)}{6}$

Can you take it from here?

--Chris

3. Hi chogz,

$\displaystyle P(n):~~~\sum^n_{r=1}{(r-1)(r+1)}=\frac{1}{6}n(n-1)(2n+5)$

Assume $\displaystyle P(k)$ is true for some $\displaystyle k \in \mathbb {N}^+$ and hence,

$\displaystyle \sum^k_{r=1}{(r-1)(r+1)}=\frac{1}{6}k(k-1)(2k+5)$

$\displaystyle \implies \sum^{k+1}_{r=1}{(r-1)(r+1)}=\frac{1}{6}k(k-1)(2k+5)+k(k+2)$

$\displaystyle = \frac{1}{6}k[(k-1)(2k+5)+6(k+2)]$

$\displaystyle = \frac{1}{6}k[2k^2+9k+7]$

$\displaystyle = \frac{1}{6}k(k+1)(2k+7)$

$\displaystyle = \frac{1}{6}(k+1)[(k+1)-1][2(k+1)+5]$

hence $\displaystyle P(k)\implies P(k+1)$ is true and thus $\displaystyle P(n)$ is true $\displaystyle \forall n \in \mathbb {Z}$ and $\displaystyle n\geq 2$

Hope this helps.

4. Thanks alot guys you've helped big time.

Really appreciate it and once again thank you.

5. Just to confirm:

<---

could you please explain why the k from where i indicated was taken out? in the next line, and to make sure the 6 came from the fraction side rite? so it be something like 6(k+2)/6 ? wouldn't it? if possible could you show the working out how you got to the next line? because i don't really fully understand it.

sorry i'm not really good with these types of question.