Question

Verify the following for n=1 & n=3
1+5+9+...+(4n-3)=n(2n-1) and prove it where n belongs to Natural numbers by mathematical induction

Answer 1

Essentially show that 1 = n(2n-1) where n=1 and 1+5+9=n(2n-1) where n=3

Answer 2

prove it by induction for k+1

Answer 3

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

Answer 4

add 4k+1 on both sides

Answer 5

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

Answer 6

2k^2-k+4k+1 <--- stuck here

Answer 7

@Xavier

Answer 8

You don't need to do anything complicated. Evaluate n(2n-1) at n=3 and show that equals 1+5+9

Answer 9

sorry but I didn't mentioned the full question

Answer 10

@Xavier ??

Answer 11

???

Answer 12

hello anyone there??

Answer 13

Ah sorry. General proof amounts to showing
n(2n-1)+(4(n+1)-3) is the same as (n+1)(2(n+1)-1)

Answer 14

yes I know but I am stuck

Answer 15

Just foil each of them out

Answer 16

what

Answer 17

Do you see how I got the two equations I listed there?

Answer 18

yes I know

Answer 19

Then expand them

Answer 20

but I have to prove reverse way

Answer 21

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

Answer 22

(4n-3)
(4(n+1)-3)
4n+1
Add 4n+1 on bs

Answer 23

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

Answer 24

2n^2-n+4k+1 <---- stuck here................
.
.
we have to make it (n+1)[2(n+1)-1]

Answer 25

This is how I think about induction.
Prove the equation holds for base case: n=1
Assume it holds for n: 1+5+9+...+(4n-3)=n(2n-1) is true
Show it holds for n+1.
1+5+9+...+(4n-3) + (4n+1)=(n+1)(2(n+1)-1)
See I substituted the n on the right with n+1. Now I know what 1+5+9+...+(4n-3) on the left is.
n(2n-1)+(4n+1) = (n+1)(2(n+1)-1). Now is this true?