Extension One Revising Game (1 Viewer)

[Timothy.Siu]

Don't quite understand the last part :S
And you were meant to type your own question!

By the way, Harder 3u is 4u

he just added the same terms to both sides

 

[Timothy.Siu]

This doesn't work.

it does

 

[youngminii]

OH, sorry I thought the (10ⁿ⁺¹ - 9n - 10) was part of the denominator. Nevermind

 

[addikaye03]

uhh ok

Prove by induction that 7+77+777+....+777...to n digits=7/81(10ⁿ⁺¹-9n-10)

G.P: last term has 7+70+700+... a=7,r=10
Sn=7/9(10^n-1)

7=S1=7/9(10-1)
70=S2=7/9(100-1)
777...=7/9(10^n-1)

BY adding n lines 7+77+777+777 n digits=7/9(10+10^2+10^3+..10^n)-7n/9
7/9 x [10(10^n-1)/9]-7n/9

=7/9x[10(10^n-1)/9-n)]
=7/81(10^n+1-9n-1)

 

[Timothy.Siu]

sorry changed it now

 

[addikaye03]

Factorise (a+b+c)^3-a^3-b^3-c^3

 

[Trebla]

Prove by induction that 7+77+777+....+777...to n digits=7(10ⁿ⁺¹-9n-10)/81

Using an induction proof...
For n = 1:
LHS = 7
RHS = 7(100 - 9 - 10) / 81
= 7
LHS = RHS so statement is true for n = 1
Assume it is true for n = k
7+77+777+....+777...to k digits = 7(10^k+1-9k-10)/81
Required to prove it is true for n = k + 1
7+77+777+....+777...to k + 1 digits = 7(10^k+2-9k-19)/81
LHS = 7+77+777+....+777...to k + 1 digits
= 7(10^k+1-9k-10)/81 + 777777.... (k + 1 digits) by assumption
= 7[10^k+1-9k-10 + 81(111111.....{k + 1 digits})] / 81
= 7[10^k+1-9k-10 + 81(10000000.....{k + 1 digits} + 10000000.... {k digits} + 10000000.... {k - 1digits} + ......)] / 81
= 7[10^k+1-9k-10 + 81(10^k + 10^{k - 1} + 10^{k - 2} + ......)] / 81
= 7[10^k+1-9k-10 + 81(10^{k + 1} - 1) / 9)] / 81
= 7[10^k+1-9k-10 + 9.10^{k + 1} - 9] / 81
= 7[10.10^k+1-9k-19)] / 81
= 7[10^k+2-9k-19)] / 81
= RHS
If the statement is true for n = k, it is also true for n = k + 1.
Since it is true for n = 1, it is true for all positive integers n by induction.

 

[Trebla]

Factorise (a+b+c)^3-a^3-b^3-c^3

(a + b + c)³ - a³ - b³ - c³ = (a + b + c - a)((a + b + c)² + a(a + b + c) + a²) - (b + c)(b² - bc + c²)
= (b + c)((a + b + c)² + a(a + b + c) + a²) - (b + c)(b² - bc + c²)
= (b + c)[(a + b + c)² + a(a + b + c) + a² - (b² - bc + c²)]
= (b + c)[a² + b² + c² + 2ab + 2ac + 2bc + a² + ab + ac + a² - b² + bc - c²]
= (b + c)[3a² + 3ab + 3ac + 3bc]
= (b + c)[3a(a + b) + 3c(a + b]
= 3(a + b)(b + c)(a + c)

I'll give a question: (It's not as straightforward as it looks)

Consider a triangle ABC with, the length of BC being 20 metres and the length of AC being 30 metres. It is also known that angle BAC is 40º.

(i) Find angle ABC to the nearest minute

(ii) Hence calculate the side AB to 2 decimal places

 

[addikaye03]

(a + b + c)³ - a³ - b³ - c³ = (a + b + c - a)((a + b + c)² + a(a + b + c) + a²) - (b + c)(b² - bc + c²)
= (b + c)((a + b + c)² + a(a + b + c) + a²) - (b + c)(b² - bc + c²)
= (b + c)[(a + b + c)² + a(a + b + c) + a² - (b² - bc + c²)]
= (b + c)[a² + b² + c² + 2ab + 2ac + 2bc + a² + ab + ac + a² - b² + bc - c²]
= (b + c)[3a² + 3ab + 3ac + 3bc]
= (b + c)[3a(a + b) + 3c(a + b]
= 3(a + b)(b + c)(a + c)

U r so VERY good at maths lol I have to admit i didnt get this Q right(stuff'd up a factorisation), u nailed it tho. have u got a Q to post up?

 

[addikaye03]

(a + b + c)³ - a³ - b³ - c³ = (a + b + c - a)((a + b + c)² + a(a + b + c) + a²) - (b + c)(b² - bc + c²)
= (b + c)((a + b + c)² + a(a + b + c) + a²) - (b + c)(b² - bc + c²)
= (b + c)[(a + b + c)² + a(a + b + c) + a² - (b² - bc + c²)]
= (b + c)[a² + b² + c² + 2ab + 2ac + 2bc + a² + ab + ac + a² - b² + bc - c²]
= (b + c)[3a² + 3ab + 3ac + 3bc]
= (b + c)[3a(a + b) + 3c(a + b]
= 3(a + b)(b + c)(a + c)

I'll give a question:

Consider a triangle ABC with, the length of BC being 20 metres and the length of AC being 30 metres. It is also known that angle BAC is 40º.

(i) Find angle ABC to the nearest minute

(ii) Hence calculate the side AB to 2 decimal places

let <ABC=@
SinB/b=SinA/a
Sin@/30=Sin40/20
Sin@=30sin40degrees/20
@=74 degrees 37'

ii) <ACB=180-(40+74.37')
=65'23', 105'23' ( nearest min)

Similarly, AB=20Sin65'23'/Sin40 degrees
=28.29, 30m (2d.p)

Question: Prove by mathematical induction, the sum of the exterior angles of a n-sided polygon=360 degrees

 

[Trebla]

let <ABC=@
SinB/b=SinA/a
Sin@/30=Sin40/20
Sin@=30sin40degrees/20
@=74 degrees 37'

ii) <ACB=180-(40+74.37')
=65'23' ( nearest min)

Similarly, AB=20Sin65'23'/Sin40 degrees
=28.29 (2d.p)

It's not that straightforward. There's something missing. I'll let you think about it...

 

[Timothy.Siu]

It's not that straightforward. There's something missing. I'll let you think about it...

uhh the other angle?
105'22'53.1'

 

[addikaye03]

It's not that straightforward. There's something missing. I'll let you think about it...

I knew it! lol it felt wayy to easy. Well im thinking.... The angle orientation allows for the above answer, there cant be an obtuse angle...its non-right angle therefore Sine rule or Cosine... lol...umm... could u please assist me lol

 

[youngminii]

What's going on? How did Timothy get his answer o.o

 

[Timothy.Siu]

well if thats all, theres no other trick, its like a yr 10 question, unless i'm missing something

 

[Trebla]

Timothy.Siu's correct. Good job. This also implies there are two possible answers for AB.

 

[addikaye03]

well if thats all, theres no other trick, its like a yr 10 question, unless i'm missing something

Angle orientation doesnt allow for obtuse, does it?

 

[youngminii]

Oh shit.
Is it because sin(180 - theta) = sin(theta)?
So theta can be 74'37' OR 105'22'?

P.S. You're a genius Trebla

 

[Timothy.Siu]

Angle orientation doesnt allow for obtuse, does it?

what?

 

[Timothy.Siu]

Oh shit.
Is it because sin(180 - theta) = sin(theta)?
So theta can be 74'37' OR 105'22'?

P.S. You're a genius Trebla

spot on