Induction problem can anyone help?? (1 Viewer)

[captainvagina]

Prove using induction that

(2/1x3) + (2/3x5) + ....... + [2/(2n-1)(2n+1)] = 1- [1/(2n+1)]


i get to a certain stage where i just cant seem to factorise any more to get the solution....

please help

 

[ninetypercent]

(2/1x3) + (2/3x5) + ....... + [2/(2n-1)(2n+1)] = 1- [1/(2n+1)]

For n =1
LHS = 2/(1)(3) = 2/3
RHS = 1 - (1/3)
= 2/3
LHS = RHS
therefore, it is true for n =1

Assume that it is true for n = k
(2/1x3) + (2/3x5) + ....... + [2/(2k-1)(2k+1)] = 1- [1/(2k+1)] - induction hypothesis

For n = k + 1
(2/1x3) + (2/3x5) + ....... + [2/(2k-1)(2k+1)] + 2/[2(k+1)-1][2(k+1)+1] = 1- [1/(2(k+1)+1)]

LHS = 1 - [1/(2k+1)] + 2/[2(k+1)-1][2(k+1)+1]
= 1 - [1/(2k+1) + 2/[(2k+1)(2k+3)]
= 1 + [(-2k-3)/[(2k+1)(2k+3)] + 2/[(2k+1)(2k+3)]]
= 1 + {(-2k-1)/[(2k+1)(2k+3)]}
= 1 - {1/[2(k+1) + 1]}
= RHS
therefore, true for n = k + 1

then u have your concluding statement

 

[captainvagina]

(2/1x3) + (2/3x5) + ....... + [2/(2n-1)(2n+1)] = 1- [1/(2n+1)]

For n =1
LHS = 2/(1)(3) = 2/3
RHS = 1 - (1/3)
= 2/3
LHS = RHS
therefore, it is true for n =1

Assume that it is true for n = k
(2/1x3) + (2/3x5) + ....... + [2/(2k-1)(2k+1)] = 1- [1/(2k+1)] - induction hypothesis

For n = k + 1
(2/1x3) + (2/3x5) + ....... + [2/(2k-1)(2k+1)] + 2/[2(k+1)-1][2(k+1)+1] = 1- [1/(2(k+1)+1)]

LHS = 1 - [1/(2k+1)] + 2/[2(k+1)-1][2(k+1)+1]
= 1 - [1/(2k+1) + 2/[(2k+1)(2k+3)]
= 1 + [(-2k-3)/[(2k+1)(2k+3)] + 2/[(2k+1)(2k+3)]]
= 1 + {(-2k-1)/[(2k+1)(2k+3)]}
= 1 - {1/[2(k+1) + 1]}
= RHS
therefore, true for n = k + 1

then u have your concluding statement

thnx heaps mate....i know why i couldnt do the question....i was using this as LHS of equation:

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

instead of:

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

lol fully misread the answer i got to n=k

 

[alcalder]

A trick you can do is use the right hand side of your page to factorise or expand the expression you are trying to get to. Then when it starts to look like what you have in your working, you can just copy the steps backwards into your answer (the right hand side of tha page stays as working, not part of your working that gets marked).

 

[captainvagina]

A trick you can do is use the right hand side of your page to factorise or expand the expression you are trying to get to. Then when it starts to look like what you have in your working, you can just copy the steps backwards into your answer (the right hand side of tha page stays as working, not part of your working that gets marked).

oh ok thnx

so wat u mean is start
RHS=...
=...
until u get to an answer

then
LHS=...
=...
until u get to that same answer?

 

[emmcyclopedia]

You can use both sides of the equation in MI.
In other words, you don't just have to manipulate the LHS.
You can use the RHS.
Or you can manipulate on side to a point, then manipulate the other to the same point. That's still showing they're the same =]