#Induction Questions
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how hard is “hard”
that doesn’t seem too bad
you get it?
you’ve proven the $\frac1{1\cdot 2}+\frac1{2 \cdot 3} + \frac1{3 \cdot 4}…$ thing right
Melon B
prove that $\frac1{1\cdot 2}+\frac1{2 \cdot 3} + \frac1{3 \cdot 4} + … = 1$
Melon B
by induction
an easier version of this that makes the induction slightly more obvious:
prove that $\frac1{1\cdot 2}+\frac1{2 \cdot 3} + \frac1{3 \cdot 4} + … + \frac1{(n-1) \cdot n} = 1 - \frac1n$
Melon B
ok cool
so the above result follows from this one, and this is a pretty classic induction q
not very hard though so idk if it’s what you want
@true tartan by any chance are you doing a level maths
year 12
anyhow theres this really good website called MadAs Maths where you can get questions for a level pure and further core maths
I also had one that’s a bit more combinatorial:
Kevin is climbing a set of n stairs, taking either one or two steps at a time. Find the number of ways he can climb these stairs.
so i reccomend going there
so which maths module did yo pick
ok
the thing with induction questions is that half the issue is bashing small cases to find the answer and the other half is actually proving it using induction
4 unit extension 2 math
wait aus??
I’ve got friends who are doing 4u
isn’t it the holidays though bro’s already grinding
locked in for exams 👺
you mean vegemite 💀 or is this some aus thing I’m too sheltered to know about
that’s crazy
it’s legit the start of the year and bro’s grinding
wish I could be that locked in
mb marmite is the british version
nah i live in uk
hell no that shi taste bland
I wish
no
i don't eat vegemite
ok so induction
same
idk about you
did you solve the first q I gave
also imagine not asking in hsc server
nearly all the ones we get in vic are trivial without induction
asuume tru for n = 0/1
what
assume true for k
wait is this still the sum of fractions
assume tru for n+k
solve it out
if tru for n =k
try this one lol idk if it’s very useful for 4u
that is use algebra
then also true for k+1
ok
let a be no. of times you climb twice, b be no. of times you climb once, let a = 1,2,3,4,5....some k where n = 2k or n = 2k+1, then arrange that many a's and (k-a) b's in a line
]
which is k choose a
2a+b = n, a,b > 0
set b = n - 2a, n = 2k or 2k+1
for each value of a, do k choose a and add them up
assuming it isn't 1 or 2 stairs
then it's 1 and 2 respectively
@true tartan
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