Mathematical Induction Question (1 Viewer)

[acevipa]

I'm having trouble with question 5 (a) of the HSC 1993 paper:

http://www.boredofstudies.org/courses/maths/3u/1095803290_1993_Mathematics_Extension_1_HSC_BOS.pdf


And am also having trouble with question 5 (a) of the HSC 1999 Paper

http://boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/99MATH34.PDF

 

[addikaye03]

Proposition: 1^2+2^2+...+n^2=1/6(n)(n+1)(2n+1)
Step 1 : Prove true for n=1

LHS=n^2=1^2=1
RHS=1/6(1)(2)(3)=1
therefore, LHS=RHS and true for n=1

Step 2: Assume true for n=k
1^2+2^2+3^2+...+k^2=1/6k(k+1)(2k+1)

Step 3: Prove true for n=k+1
1^2+2^2+3^2+...+k^2+(k+1)^2=1/6(k+1)(k+2)(2k+3)

LHS (through sub of step 2)=1/6k(k+1)(2k+1)+(k+1)^2
=(k+1)((1/6k(2k+1)+(k+1))
by multiplying by 6,=1/6(k+1)(k(2k+1)+6(K+1))
=1/6(k+1)(k+2)(2k+3)..... Bracket was expanded and quadratic was factorised
=RHS

Since true for n=1 and n=k+1, then true for n=2, and n=3 and so on for all positive integar to the value of n

Hope this helps, it was a prick to write out lol

 

[acevipa]

Part ii was what I was mainly having trouble with

 

[proringz]

If Sn >= 10^9 then
1/6n(n+1)(2n+1) >= 10^9 i.e n(n+1)(2n+1) >= 6 x 10^9

Now 6x10^9 very large, therefore n must be very large.
As n--> infinity, (n+1) --> n, and (2n+1) --> 2n
Thus, least value for n;
n(n+1)(2n+1) >= 6x10^9
From above,
n x n x 2n = (approx) 6 x 10^9
i.e 2n^3 = 6 x 10^9
n^3 = 3 x 10^9
i.e n = cubedroot(3 x 10^9)
= 1442 nearest integer

It'll look much simpler if I wrote it down.

 

[addikaye03]

Part ii was what I was mainly having trouble with

sorry, i didnt see that part.. ok

1/6n(n+1)(2n+1)>=10^9
1/6(n^2+n)(2n+1)>=10^9
2n^3+3n^2+n>=6(10^9)
n^3+3/2n^2+1/2n>=3(10^9)

For large numbers n and n^2 become irrelevant in comparison to n^3
n^3=3(10^9) (approx)
therefore n=1442

u should prob sub in n=1441 and 1442 to show that 1441 is <3(10^9) and > 3(10^9) respectively.

 

[addikaye03]

If Sn >= 10^9 then
1/6n(n+1)(2n+1) >= 10^9 i.e n(n+1)(2n+1) >= 6 x 10^9

Now 6x10^9 very large, therefore n must be very large.
As n--> infinity, (n+1) --> n, and (2n+1) --> 2n
Thus, least value for n;
n(n+1)(2n+1) >= 6x10^9
From above,
n x n x 2n = (approx) 6 x 10^9
i.e 2n^3 = 6 x 10^9
n^3 = 3 x 10^9
i.e n = cubedroot(3 x 10^9)
= 1442 nearest integer

It'll look much simpler if I wrote it down.

beat me to it lol