Re: Prove by mathematical induction. (1 Viewer)

[EViL.GENiUS]

Hi could someone help me here.

Prove by mathematical induction.

 

[Trebla]

When n = 1
LHS = 1 + 1 = 2
RHS = 1(3 + 1)/2 = 2
LHS = RHS
.: statement is true for n = 1
Assume the statement is true for n = k
(k + 1) + (k + 2) + ...... + (2k - 1) + (2k) = k(3k + 1)/2
Need to prove it is true for n = k + 1
(k + 2) + (k + 3) + ...... + (2k + 1) + (2k + 2) = (k + 1)(3k + 4)/2
LHS = (k + 2) + (k + 3) + ...... + (2k) + (2k + 1) + (2k + 2)
= k(3k + 1)/2 - (k + 1) + 2k + 1 +2k + 2 by assumption
= k(3k + 1)/2 + 3k + 2
= (3k² + 7k + 4)/2
= (k + 1)(3k + 4)/2
= RHS
[insert conclusion here....]

 

[EViL.GENiUS]

from line 12&13

= (3k² + 5k + 4)/2
= (k + 1)(3k + 4)/2

3k² + 5k + 4 doesn't equal (k + 1)(3k + 4)

 

[donetha]

are you sure there isn't a mistake in the question, because I am getting the same second last line as the other guy...(3k^2+5k+4)/2..but that doesn't factorise nicely.

I can get to (3k+2)(k+1)/2 but not (3k+4)(k+1)/2

 

[EViL.GENiUS]

are you sure there isn't a mistake in the question,

I checked the question again it is correct.

 

[donetha]

I checked the question again it is correct.

hmmm....maybe we're missing something then..

 

[3unitz]

Hi could someone help me here.

Prove by mathematical induction.

assume true for n = k,
(k + 1) + (k + 2) + ... + (2k)= k(3k + 1)/2

n = k+1
[(k + 1) + 1] + [(k + 1) + 2] + ... + [(k+1) + (k+1)] = (k+ 1) [3(k + 1) + 1]/2

LHS = [(k + 1) + 1] + [(k + 1) + 2] + ... +[(k+1) + k] + [(k+1) + (k+1)]
= k(3k + 1)/2 + k + [(k+1) + (k+1)] (using assumption)***
= k(3k + 1)/2 + 2k/2 + 4(k+1)/2
= (3k^2 + k + 2k + 4k + 4)/2
= (3k^2 + 7k + 4)/2
= (k+1)(3k+4)/2

RHS = (k+ 1) [3(k + 1) + 1]/2 = (k+1)(3k+4)/2 = LHS

*** the colour parts just try to show how the assumption was used:
the red parts is the assumption
the blue parts are the left over 1's (which add to give k)
the orange part is just the left over part

 

[3unitz]

are you sure there isn't a mistake in the question

you can always check the question yourself using series:

LHS = (n + 1) + (n + 2) + (n + 3) + ... + (n + n)
= (n x n) + (1 + 2 + 3 + ... + n)
= n^2 + (n/2)(1 + n)
= (2n^2 + n + n^2)/2
= (3n^2 + n)/2
= n(3n + 1)/2
= RHS

 

[bobos2]

before my teacher taught us induction, she said it was really really difficult, and but i've learnt it, && i've found it pretti easy...is there something im missing? did u guys find it that difficult to understand? thanx.

 

[EViL.GENiUS]

Induction questions vary in difficulty but mostly it just, put n=1, the n=k and try to make n=k+1 from n=k

 

[tommykins]

What the fuck at your sig.