Mathematical Induction Question (1 Viewer)

[pakigal]

Hey guys I've got this question which seems alright but for some reason I'm not getting the answer! grrr!! It's really pissing me off now cos I know I'm doing something stupid, can someone please try do it for me. I wud scan my answer except I don't have a scanner...
ok so the question is:

Prove by mathematical Induction that:

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

Thanks

 

[Trebla]

I'm assuming you have problems with the n = k + 1 step:

LHS = 2/(1x3) + 2/(3x5) + 2/(5x7) + ... + 2/(2k-1)(2k+1) + 2/(2k+1)(2k+3)
= 1 - 1 / (2k+1) + 2 / (2k+1)(2k+3) by assumption

We can reduce the sign confusion by expanding the minus sign into the numerator: (I'm guessing this is where the error may have occured - which is a very common one)

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

 

[pakigal]

Yeah that's where it occurred. I went over it 5 times but couldn't figure out what I was doing wrong.
Thanks

 

[watatank]

Prove by mathematical Induction that:

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

Thanks

for n = 1

LHS = 2/3
RHS = 1-1/3 = 2/3

assume for some k

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

then for n = k+1

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


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

your equation goes to
1 - 1/(2k+1) + 2/(2k+1)(2k+3) = 1 - 1/(2(k+1)+1)

or

1 - 1/(2k+1) + 2/(2k+1)(2k+3) = 1 - 1/(2k+3)

1's cancel out

- 1/(2k+1) + 2/(2k+1)(2k+3) = - 1/(2k+3)

put LHS over a common denominator.

[-(2k+3)+ 2] / [ (2k+1)(2k+3) ] = - 1/(2k+3)

[-(2k+1)] / [ (2k+1)(2k+3) ] = - 1/(2k+3)

cancel (2k+1) out of num and denom of LHS

and youve got the LHS = RHS

so by mathematical induction, true for all real n

dodgy working out, but it works! :wave:

 

[pakigal]

Thanks pplz! Gosh, I really don't like this topic! Can you help me with this question as well?

Prove the following by finite (mathematical) induction for integers greater than or equal to 1:

If U(n+1)= 2Un + 1 and U1=1, then Un= (2^n) - 1

Thanks

 

[ellen.louise]

Thanks pplz! Gosh, I really don't like this topic! Can you help me with this question as well?

Prove the following by finite (mathematical) induction for integers greater than or equal to 1:

If U(n+1)= 2Un + 1 and U1=1, then Un= (2^n) - 1

Thanks

Prove true for n=1

U1 = 2^1 - 1
U1 = 1
therefore true for n=1

assume true for n=k, (i.e. Uk = 2^k - 1)
Prove true for n=k+1 (i.e. Prove U(k+1) = 2^(k+1) -1)

AND:
U(n+1) = 2Un + 1


LHS
= U(k+1)
=2Uk + 1
=2(2^k - 1) + 1
=2.2^k - 2 + 1
=2^(k+1) - 1
=RHS