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oh wow I just saw yours. Can you do that to every subject and if you remember how does it cost for one

You can do that? wow

How do you even remember the exact essay u wrote in the HSC? Was it a pre planned essay you memorised because your exaggerating right?

I did a billion times mate finally works now tho

I dont see anything... its blank for me

Is part a fine tho?

Oh nvm it did sorry

It doesnt pop up I see (from part A) then raising both sides to or that

???

Can someone check this Im not sure for part A : For part b - I literally cant do it Idk how to approach it A) Use mathematical Induction to prove that 3^n >n^3 for all integers n >=4 For n=4 LHS= 81 RHS= 64 Therefore LHS > RHS Thus true Assume true for n=k, k all real 3^k > k^3 RTP; n=k+1...

so they pulled it out of thin air just to make the inequality hold? coz if I did this myself I would never ever think of doing that

How did they go from = k^3 - (3k^2 +3k+1) to >k^3 - (3k^2 +3k^2+3k^2) ??????

inequality and geometric induction is such a bother for me ughh is that part of 3u? heard its in yr 12 and Im only starting yr 11 this year so idk

yeah those are ez

Talking about mathematical induction in general does it appear much? Fingers crossed it doesnt 😭

ughh I hate this imao how important is this topic in 3u and 4u? Like does it appear much in exams etc

THANK U SO MUCHH and btw how would you prove by MI that the sum of exterior angles of an n sided convex polygon is 360 like what do I say for n=k and n=k+1

Everytime I do that question above I end up getting a negative answer for a >= 0. I think im doing the question completely wrong

Im struggling with this question: Prove by MI that 2^2^n >= 5^2n for all integers n>=5 I did: For n=5, LHS=42967296 RHS=9765626 Therefore LHS>=RHS Thus true Assume true for n=k, 2^2^k >= 5^2k RTP; n=k+1 2^2^k+1 >= 5^2k+2 LHS= 2^2^k . 2^2 >= 4(5^2k) - from assumption RHS- LHS=...