## anonymous one year ago MATHEMATICAL INDUCTION

1. anonymous

$1^{2}+4^{2}+7^{2}+....+(3n-2)^{2}=\frac{ n(6n ^{2}-3n-1) }{ 2 }$

2. anonymous

How would I do s(k+1)? @ganeshie8

3. IrishBoy123
4. anonymous

Okay thanks that shows me the work but I dont understand it, can you explain to me step by step? @IrishBoy123

5. IrishBoy123

sure in inductions, you first prove it is true for some value of n. so establish that bit first.

6. anonymous

I already did that part: n=1 and n=k I just dont know how to d the n=k+1

7. anonymous

Honestly, I don't get it.

8. IrishBoy123

OK, one important thing. last time i helped on this the OP specified "for all positive integers n" so yes you had to start with n = 1. we will use induction to prove that it must therefore be true for n = 2, and thuse n = 3 and thus .... so the next step is to assume it is true for n = k. that means the following is assumed true |dw:1440079974650:dw|

9. anonymous

Yeah I have that

10. IrishBoy123

i follows therefore that : |dw:1440080053278:dw| that has to be true as we have added the same amount to each side as it happens what we added to LHS was next term in the series

11. IrishBoy123

and the critical step, do you know that too?

12. anonymous

that step just confused me im trying to understand it just a min

13. anonymous

Question: is it not that for n =k, we have $$....+(3k-2)^2= \dfrac{k(6k^2-3k-1)}{2}$$

14. anonymous

|dw:1440080370055:dw|

15. anonymous

Woah im so lost now

16. IrishBoy123

|dw:1440080425206:dw|

17. anonymous

oh okay so 3(k+1-2)^2 takes the place of 6k^2....

18. anonymous

yeah

19. IrishBoy123

slow down a bit gives me a minute to summarise where we are: will take less than a minute

20. IrishBoy123

|dw:1440080585980:dw|

21. IrishBoy123

|dw:1440080618737:dw|

22. IrishBoy123

so the final step is to show that the RHS is as the original equation would predict ie |dw:1440080671229:dw|

23. IrishBoy123

and that's proof done, if we complete that final step

24. IrishBoy123

so we now wrestle with the algebra trying to show these 2 are in fact equal.

25. anonymous

I did, they are not the same :(

26. anonymous

I show you

27. IrishBoy123

28. anonymous

Alright I have to go ill be back though sorry You guys can continue helping crazyand... I will catch up with you guys

29. IrishBoy123

that's all you need to know, the rest is the algebra. it's turgid, that's why i shared my working -- which shows one approach

30. anonymous

$$\dfrac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2\\=\dfrac{k(6k^2-3k-1)}{2}+(3k+1)^2\\=\dfrac{k(6k^2-3k-1)}{2}+9k^2+6k+1\\=\dfrac{k(6k^2-3k-1)+18k^2+12k+2)}{2}\\$$ $$=\dfrac{6k^3-3k^2-k+18k^2+12k+2}{2}$$ $$=\dfrac{6k^3+15k^2+11k+2}{2}$$

31. IrishBoy123

right, so far. at this point i divided by k + 1 as we know that is on the RHS if the statement is correct

32. anonymous

while other is $$\dfrac{(k+1)(6(k+1)^2-3(k+1)-1}{2}$$ $$=\dfrac{(k+1)(6(k^2+2k+1))-3k-3-1}{2}$$ $$\dfrac{(k+1)(6k^2+12k+6)-3k-4}{2}$$ $$=\dfrac{6k^3+12k^2+6k+6k^2+12k+6-3k-4}{2}$$ $$=\dfrac{6k^3+18k^2+15k+2}{2}$$ Which is different from the previous one. :(

33. IrishBoy123

check brackets

34. anonymous

YYYYYYYYYYYES!! I got you. Finally. Thanks for being patient to me.

35. IrishBoy123

thank you for showing me a shorter proof. i don't know what the hell i was doing using long division last time i did this with someone .... just plain nuts.

36. anonymous

We learned from each other, right? hehehe..