how do you do these plz help (2 Viewers)

[v.tex]

plz send working

 

[Masaken]

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plz send working

i might be cooked since it's been a very long time since 4u hsc for me but you can use mathematical induction to solve both. but for the first q the statement just isn't true because for the base case n=1 => 3 + 2 = 5 which is perfectly divisible by 5 with no remainder so you can't prove it. same for n=2 wherein 3^2 + 2^2 = 13, when divided by 5 that's a remainder of 3

second q is mathematical induction if i recall correctly

• base case: where n = 1 => 5^(1) - 1 = 4 which is divisible by 4
• assume true for n = k i.e. 5^k - 1 = 4Q (wherein Q is a positive integer)
• prove true for n = k+1 i.e. 5^(k+1) - 1 = 4T (wherein T is positive integer)
  • basically i don't wanna send the full working out here cos like idk how to format it so it'll look kinda ugly but by index laws break up 5^(k+1) into 5^k * 5^1 and then from the assumption of n = k; rearranging: 5^k = 4Q + 1, sub that into the equation you're trying to prove true and you should be able to prove it true and thus make the conclusion

 

[WeiWeiMan]

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plz send working

4. maybe consider 3 = 5-2 then binomial, tho idk if that give remainder of 0, thus divisible by 5 (including base case)

1. x^n-1 = (x-1)(1+x+x^2 +…+x^n-1)

 

[WeiWeiMan]

i might be cooked since it's been a very long time since 4u hsc for me but you can use mathematical induction to solve both. but for the first q the statement just isn't true because for the base case n=1 => 3 + 2 = 5 which is perfectly divisible by 5 with no remainder so you can't prove it. same for n=2 wherein 3^2 + 2^2 = 13, when divided by 5 that's a remainder of 3

second q is mathematical induction if i recall correctly

• base case: where n = 1 => 5^(1) - 1 = 4 which is divisible by 4
• assume true for n = k i.e. 5^k - 1 = 4Q (wherein Q is a positive integer)
• prove true for n = k+1 i.e. 5^(k+1) - 1 = 4T (wherein T is positive integer)
  • basically i don't wanna send the full working out here cos like idk how to format it so it'll look kinda ugly but by index laws break up 5^(k+1) into 5^k * 5^1 and then from the assumption of n = k; rearranging: 5^k = 4Q + 1, sub that into the equation you're trying to prove true and you should be able to prove it true and thus make the conclusion

Yeah I’m lowkey cooked I forgot to check numbers

the question should be divisible by 5 for odd integers n, as the -2^n from the binomial only exists (and thus cancels with 2^n) when n is odd, otherwise it’d be +2^n

anyway that should be the easiest way to prove it since yk (-1)^odd = -1

 

[eunoia]

for the second question; this is how I usually set out my inductions
.