Mathematical Induction Q's (1 Viewer)

[puffynutty]

I got a few past hsc questions which i have been doing, but i don't know if i've been doing them correctly as i don't have the answers.

1. Use the principle of mathematical induction to show that

2x1!+5x2!+10x3!+…+(n²+1)n!=n(n+1)!

For all integers positive n.



2.Use Mathematical induction to prove that

1/(1x3) + 1/(3x5) + 1/(5x7) +...+ 1/(2n-1)(2n+1)=n/(2n+1)

For all integers positive n.

 

[tommykins]

I'll do it fully once hten I'll skip the n = 1 steps.

1. Use the principle of mathematical induction to show that

2x1!+5x2!+10x3!+…+(n²+1)n!=n(n+1)!

For all integers positive n.

Prove true for n = 1
((1)²+1)*1! = 2*1! = 2
1(1+1)! = 1*(2!) = 2

LHS = RHS .'. true for n = 1

Assume true for n = k
That is,
2x1!+5x2!+10x3!+…+(k²+1)k!=k(k+1)!

Prove true for n = k + 1
That is,
2x1!+5x2!+10x3!+…+(k²+1)k!+[(k+1)²+1](k+1)!=(k+1)(k+2)!
LHS = k(k+1)!+[(k+1)²+1](k+1)! -> from the assumption
= (k+1)![ k + (k+1)² +1 ]
= (k+1)![ k + k² + 2k + 1 + 1 ]
= (k+1)![ k² + 3k + 2]
= (k+1)!(k+1)(k+2)
= (k+2)!(k+1)
= RHS

.'. By principle of mathematical induction it is true for all n > 1

2.Use Mathematical induction to prove that

1/(1x3) + 1/(3x5) + 1/(5x7) +...+ 1/(2n-1)(2n+1)=n/(2n+1)

For all integers positive n.[/SIZE]
[/SIZE]

I'll go straight into it .

n = k
1/(1x3) + 1/(3x5) + 1/(5x7) +...+ 1/(2k-1)(2k+1)=k/(2k+1)

prove for n = k+1
1/(1x3) + 1/(3x5) + 1/(5x7) +...+ 1/(2k-1)(2k+1) + 1/(2k+1)(2k+3) = (k+1)/(2k+3)

LHS = k/(2k+1) + 1/(2k+1)(2k+3)
= [k(2k+3) + 1]/(2k+1)(2k+3)
= [ 2k²+3k+1]/(2k+1)(2k+3)
= (2k+1)(k+1)/(2k+1)(2k+3)
= (k+1)/(2k+3)
= RHS

PMI .

 

[puffynutty]

i like you so much your like mathematical dictionary, thanks heaps

 

[Timothy.Siu]

i like you so much your like mathematical dictionary, thanks heaps

yeah he's on a role today

 

[tommykins]

its okay, keep 'em coming.

 

[Iruka]

You know the proper way to do the 2nd Q don't you? (I mean, stuff what the question actually asks for... I hate mathematical induction - especially when you don't have to use it.)

You split up 1/(2n-1)(2n+1) using partial fractions and make a telescoping sum.

 

[Trebla]

Is that the same term as a 'collapsing sum'?

 

[Iruka]

I think so.