# Mathematical Induction Problem (IB Math HL)

• Jun 19th 2012, 09:06 AM
alireza1992
Mathematical Induction Problem (IB Math HL)
Hi. There was a question in my math book which I couldn't solve. And I would appreciate if anyone could give me the solution.
The question is about mathematical induction and it's from IB math HL textbook.
Bellow is the question (It's in the picture)
Attachment 24108
• Jun 19th 2012, 10:57 AM
richard1234
Re: Mathematical Induction Problem (IB Math HL)
The base cases n = 1, 2, 3 can be verified to be true. Assume it is true for some n = k. We want to show that

$(k+1) + 2k + 3(k-1) + ... + (k-1)3 + k(2) + (k+1) = \frac{(k+1)(k+2)(k+3)}{6}$

Using our useful "hint," rewrite the LHS as

$k + 2(k-1) + 3(k-2) + ... + (k-1)2 + k + (1+2+...+k+(k+1))$

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

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

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

$=\frac{(k+1)(k+2)(k+3)}{6}$
• Jun 19th 2012, 11:09 AM
alireza1992
Re: Mathematical Induction Problem (IB Math HL)
Thanks a a lot :)
Just one question, how did you get
(k+1)(k+2)/2
from (1+2+...+k+(k+1))
I couldn't follow that part. Could you please explain it a bit?
And thank you for helping me.
• Jun 19th 2012, 11:13 AM
richard1234
Re: Mathematical Induction Problem (IB Math HL)
The sum of the integers from $1$ to $a$ is $\frac{a(a+1)}{2}$ (which can also be proved by induction!)

Therefore the sum of the integers from $1$ to $k+1$ is $\frac{(k+1)(k+2)}{2}$.
• Jun 19th 2012, 11:16 AM
alireza1992
Re: Mathematical Induction Problem (IB Math HL)
I see now. Thank you so much! :)
• Jun 19th 2012, 11:17 AM
richard1234
Re: Mathematical Induction Problem (IB Math HL)
No problem :)
• Jan 31st 2013, 12:31 PM
DEEMATHWHIZ
Re: Mathematical Induction Problem (IB Math HL)
How did you get from k+2(k-1) +3(k-2)..... +(k-1)2+k
(.....
resulting k(k+1)(k=2)/6 ?