Induction HSC Question (1 Viewer)

[acevipa]

I'm having trouble with question 5 (a) of the HSC 1999 Paper

http://boardofstudies.nsw.edu.au/hsc...s/99MATH34.PDF

 

[5647382910]

gee i cant do it either, wtf
could someone please help

 

[adnan91]

When u wanna step n=k+1 DONT write the value of n=k plus n=k+1. Just leave n=k+1 substituted alone for LHS and RHS of the original.

 

[Trebla]

For n = 1
LHS = (1 + 1) = 2
RHS = 2¹ = 2
LHS=RHS hence statement is true for n = 1
Assume the statement is true for n = k
(k + 1)(k + 2).....(2k - 1)2k = 2^k[1 x 3 x ......... x (2k - 1)]
Required to prove statement is true for n = k + 1
(k + 2)(k + 3).....(2k + 1)(2k + 2) = 2^{k + 1}[1 x 3 x ......... x (2k + 1)]
LHS = (k + 2)(k + 3).....(2k + 1)(2k + 2)
= 2(k + 1)(k + 2)(k + 3).....2k(2k + 1)
= 2(2k + 1)2^k[1 x 3 x ......... x (2k - 1)] by assumption
= 2^{k + 1}[1 x 3 x ......... x (2k - 1) x (2k + 1)]
= RHS
If the statement is true for n = k, it is true for n = k + 1
Since the statement is true for n = 1, it is true for all positive integers n by induction.

 

[acevipa]

Thanks Trebla for the help, I think I understand it. Though could I just ask you one thing. How did you get

For n = 1
LHS = (1 + 1) = 2
RHS = 21 = 2

 

[Trebla]

Thanks Trebla for the help, I think I understand it. Though could I just ask you one thing. How did you get

For n = 1
LHS = (1 + 1) = 2
RHS = 21 = 2

(n + 1)(n + 2).....(2n - 1)2n = 2ⁿ[1 x 3 x ......... x (2n - 1)]
When n = 1, the LHS is just the first bracket (n + 1) = (1 + 1) = 2 and the RHS extracts the first multiple in the brackets giving 2¹ x 1 = 2.

 

[acevipa]

Okay. Thanks, I understand now.