## Pssssst Group Title SOMEONE PLEASE HELP I'M ALMOST DONE. Use mathematical induction to prove the statement is true for all positive integers n. 8 + 16 + 24 + ... + 8n = 4n(n + 1) one year ago one year ago

1. Pssssst Group Title

@ParthKohli

2. ParthKohli Group Title

True for the base case. Now assume that it is true for k, and then prove for k + 1.

3. ParthKohli Group Title

Can you find the relationship between those two?

4. Spacelimbus Group Title

Is it supposed to be $\Large 4n(n+1)$ If that's the case, it must run a bit strangely. $n=1 \longrightarrow 8 \checkmark \\n=2 \longrightarrow24 \neg\checkmark$

5. ParthKohli Group Title

Actually, the order of operations first tell you to multiply 4 and $$n$$, then $$n +1$$

6. Spacelimbus Group Title

I believe I can't follow you @ParthKohli $\Large 4n(n+1)=4n^2+4n$ right? and still n=1 -> 8 n=2 -> 24, or where do I miss something? (which is indeed possible=

7. ParthKohli Group Title

The question is actually saying $$(4 \cdot n) \cdot (n + 1)$$

8. ParthKohli Group Title

But 8 + 16 = 24

9. ParthKohli Group Title

It is a series, not a sequence.

10. Spacelimbus Group Title

oh I understand, this is where I misunderstood the problem, thank you.

11. ParthKohli Group Title

No problem!

12. Spacelimbus Group Title

It's not explicit mentioned in the problem set, I have seen such problems as mainly sequences/progressions, in which this would be arithmetic.

13. ParthKohli Group Title

Spacelimbus, can you please help the asker? I wish I could, but I am currently on a mobile device and it is hard for me to type.

14. tcarroll010 Group Title

Now that you know it works for the base case, you assume it holds for the case of "k", and you try to prove it works for "k + 1". So, add 8(k + 1) to both sides: 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4k(k + 1) + 8(k + 1) 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4k^2 + 4k + 8k + 8 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4k^2 + 12k + 8 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4(k^2 + 3k + 2) 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4(k + 1)(k + 2) 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4(k + 1)[(k + 1) + 1] But that is just the same expression we assumed true, but now holding for "k + 1", so this proves the case.

15. ParthKohli Group Title

^ that is simple induction, correct.

16. tcarroll010 Group Title

All good now @Pssssst ?

17. tcarroll010 Group Title

The reason this proves the case is because of the way mathematical induction works. The base case has already been shown to be true. Our first "k" is "1" and we have proved it for "k + 1" or "2". "2" becomes our new "k", and by induction, it is proved for "k + 1" or "3", and this process goes on indefinitely. That is why mathematical induction works and why the above proof works.

18. Pssssst Group Title

Thank you, everyone! Helped a lot :)

19. tcarroll010 Group Title

Glad we were able to help. Thanks for the recognition!

20. Pssssst Group Title

@tcarroll010 Do you know much about history?