Mathematica Proof of 4n-3 = n(2n-1) via Mathematical Induction

[lemurs]

ok I am really confused now topic says it all..

I am given 4n-3 = n(2n-1)

using mathemadical induction proof that is true.

P(1) both equal 1

P(k) 4k-3 = k(2k-1)
= k^2 - k

P(k+1) 4(k+1)-3 =(k+1)(2(k+1)-1)

if i simplify it all i get that
4k +1=2k^2 +3k +1

but stick at that point.

any help please.

 

[quasar987]

The proposition 4n-3 = n(2n-1) for all natural numbers is false; no wonder you can't prove it. Take n=5 for exemple. It would then say that 17=45

 

[semc]

hmm...did u miss out the summation sign on the left side?

 

[lemurs]

kay here is the exact question from the text.

1+5+9+...+(4n-3)=n(2n-1)

so how do i do this then...

 

[neutrino]

1+5+9+...+(4n-3)=n(2n-1)

You do realize that it's not the same as what you said in the first post?

Assuming it's true for some k, add 4(k+1)-3 to the left and try to simplify it so that you get the corresponding term for k+1 on the right.

 

[semc]

kay here is the exact question from the text.

1+5+9+...+(4n-3)=n(2n-1)

so how do i do this then...

(4n-3) is not summation (4n-3) sub n=k+1 on the right and proof that its equal to k(2k-1) + (k+1)th term. i guess it should be alrite from here