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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

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

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

Using our useful "hint," rewrite the LHS as

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

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

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

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

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

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.

Re: Mathematical Induction Problem (IB Math HL)

The sum of the integers from $\displaystyle 1$ to $\displaystyle a$ is $\displaystyle \frac{a(a+1)}{2}$ (which can also be proved by induction!)

Therefore the sum of the integers from $\displaystyle 1$ to $\displaystyle k+1$ is $\displaystyle \frac{(k+1)(k+2)}{2}$.

Re: Mathematical Induction Problem (IB Math HL)

I see now. Thank you so much! :)

Re: Mathematical Induction Problem (IB Math HL)

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 ?