Solutions required for following HSC Questions (1 Viewer)

[HSCstudent08]

Would you guys be kind enough to scan/upload/post the solutions to particular questions of Maths EXT 1 papers? thanks Solutions needed for: 1997 Question 5 b) 1993 Question 6 C) these are what i need atm...

 

[kelllly]

1997 Question 5 b)

(b)(i) LHS = nCr + nC(r+1) = n! / (n-r)!r! + n! / (n-r-1)!(r+1)! which is n! / (n-r)(n-r-1)!r! + n! / (n-r-1)!(r+1)r! = n! / (n-r-1)!r! [ (1 / n-r) + (1 / r+1) ] = n! / (n-r-1)!r! x (r+1+n-r) / (n-r)(r+1) = n! / (n-r-1)!r! x (n+1) / (n-r)(r+1) = (n+1)! / (n-r)!(r+1)! which is (n+1)! / (n+1-r-1)!(r+1)! = (n+1)C(r+1) = RHS (b)(ii) Step 2: Assume: kSUM(j=3) (j-1)C2 = kC3 Step 3: Prove: (k+1)SUM(j=3) (j-1)C2 = (k+1)C3 (k+1)SUM(j=3) (j-1)C2 = kSUM(j=3) (j-1)C2 + [(k+1)-1]C2 = kC3 + kC2 from assumption = (k+1)C3 from part (i)

 

[kelllly]

*sigh* at the above. I'm sorry HSCstudent08 but BOS isn't functioning normally for me.

 

[HSCstudent08]

thanks i can do that part it's just part 2

 

[kelllly]

Assume: kSUM(j=3) (j-1)C2 = kC3

 

[kelllly]

Prove: (k+1)SUM(j=3) (j-1)C2 = (k+1)C3

 

[kelllly]

(k+1)SUM(j=3) (j-1)C2

 

[kelllly]

= kSUM(j=3) (j-1)C2 + [(k+1)-1]C2

 

[kelllly]

= kC3 + kC2 from assumption

 

[kelllly]

= (k+1)C3 from part (i)

 

[untouchablecuz]

use < br > without the spaces to seperate the lines

 

[kelllly]

Oh, thanks!

 

[HSCstudent08]

ok new question: 1997 7 b) part i

 

[jet]

 

[HSCstudent08]

thanks mate!