Induction [math question]
Hi guys just wondering if I can get a little help with a math question,
lets say we know that 4n - 1 = n(2n + 1)
first we need to solve for the base case : this is simple all we do is sub 1 in for n and by doing the math we do indeed get 1 = 1 , so the base case holds true
next we need to do the induction step
so let k = n : k = k(2n + 1)
k+1:
so we know that k = k(2n + 1), therefore k+1 = k(2n + 1) + 4k - 1
so I get this part the sum before k+1 which is k is obviously equal to k(2n + 1)
but lets say we are solving for 1 we just sub 1 in for n, lets say we are solving for k we would just replace n with k so
k = 4(k) - 1 = k(2(k) + 1
so why don't we do the same for k+1 ??
why don't we just do 4(k+1) = k+1( 2(k+1) +1 )?
instead we put k ( 2(k) + 1 ) + 4(k + 1) = k+1( 2(k+1) +1 )
why do we need to k before 4( k + 1)?
wouldn't 4(k + 1) not = k+1( 2(k+1) +1 ) ??
note I did the equation and k ( 2(k) + 1 ) + 4(k + 1) does = k+1( 2(k+1) +1 ) but I'm still wondering why we need the inclusion of k on the left hand of our expression
video I'm following - https://www.youtube.com/watch?v=tHNVX3e9zd0 - first example in the video
hope that makes sense,
thanks
lets say we know that 4n - 1 = n(2n + 1)
first we need to solve for the base case : this is simple all we do is sub 1 in for n and by doing the math we do indeed get 1 = 1 , so the base case holds true
next we need to do the induction step
so let k = n : k = k(2n + 1)
k+1:
so we know that k = k(2n + 1), therefore k+1 = k(2n + 1) + 4k - 1
so I get this part the sum before k+1 which is k is obviously equal to k(2n + 1)
but lets say we are solving for 1 we just sub 1 in for n, lets say we are solving for k we would just replace n with k so
k = 4(k) - 1 = k(2(k) + 1
so why don't we do the same for k+1 ??
why don't we just do 4(k+1) = k+1( 2(k+1) +1 )?
instead we put k ( 2(k) + 1 ) + 4(k + 1) = k+1( 2(k+1) +1 )
why do we need to k before 4( k + 1)?
wouldn't 4(k + 1) not = k+1( 2(k+1) +1 ) ??
note I did the equation and k ( 2(k) + 1 ) + 4(k + 1) does = k+1( 2(k+1) +1 ) but I'm still wondering why we need the inclusion of k on the left hand of our expression
video I'm following - https://www.youtube.com/watch?v=tHNVX3e9zd0 - first example in the video
hope that makes sense,
thanks
Last edited on
Adam2016,
If
4n-1=n(2n+1)
then
n=1
or
n=1/2
and that's it. There is no induction there: it's not an identity.
If
4n-1=n(2n+1)
then
n=1
or
n=1/2
and that's it. There is no induction there: it's not an identity.
Oh, wow. That's some seriously sloppy notation. He didn't mean
4 * n - 1 = n * (2 * n + 1)
He meant
sum_{i = 1}^{n} (4 * i - 1) = n * (2 * n + 1)
4 * n - 1 = n * (2 * n + 1)
He meant
sum_{i = 1}^{n} (4 * i - 1) = n * (2 * n + 1)
Last edited on
GIVEN:
A series An = {3, 7, … , (4n -1)}
TO PROVE:
The sum of the series Sn = 1 + 3 + 7 + … + (4n -1) = n(2n+1)
PROOF BY INDUCTION:
a) For n = 1 //<- not n = 3 which was a blooper
An = {3}
LHS: Sn = 3
RHS: Sn = 1(2x1 + 1)
b) For n = k
An = {3, 7, … , (4k-1)}
LHS: Sn = 1 + 3 + 7 + … + (4k -1)
Assume RHS: Sn = k(2k + 1) is true
c) Test for n = k +1
An = {3, 7, … , (4k-1), ( 4(k+1) – 1 )}
LHS: Sn = 1 + 3 + 7 + … + (4k -1) + ( 4(k+1) – 1 )
RHS: Sn = (k+1)( 2(k+1) +1 )
So,
LHS becomes k(2k +1) + ( 4(k+1) – 1 ) = 2k2 + 5k +3
RHS becomes (2(k2 +2k +1) + k + 1) = 2k2 + 5k + 3 //<-- brackets blooper fixed
LHS = RHS
QED |
Edit: Restored equilibrium with correction to a)
Last edited on
Ah, I see! You meant summation ... without putting a summation symbol!
It's quicker as an arithmetic series:
Base case: When n=1, LHS = RHS so true by verification.
If true for a particular n, then
Sn = n(2n+1), where Sn is sum to n terms.
Sn+1 = Sn + (n+1)th term
= n(2n+1) + (4(n+1) - 1)
= 2n2 + 5n + 3
= (n+1)(2n+3)
= [n+1] ( 2[n+1] + 1 )
so true for n+1 also. |
It's quicker as an arithmetic series:
Sn = (number of terms) x (average term) = n x ( first + last ) / 2 = n x ( 3 + 4n-1 ) / 2 = n ( 4n + 2 ) / 2 = n ( 2n + 1 ) |
Yeah horrible misinterpretation on my part, it's just a summation ( which I forgot to include in the example )
that's why k* (( 2 * k) + 1) + ( 4 * k+1 ) - 1 = k+1( 2(k+1) + 1 )
reason being we need the summation of the previous numbers before we can add our new result to that summation hence why we need k* (( 2 * k) + 1) on the right hand side before ( 4 * k+1 ) - 1 .
serious blunder on my part, I don't know what I was thinking :/
that's why k* (( 2 * k) + 1) + ( 4 * k+1 ) - 1 = k+1( 2(k+1) + 1 )
reason being we need the summation of the previous numbers before we can add our new result to that summation hence why we need k* (( 2 * k) + 1) on the right hand side before ( 4 * k+1 ) - 1 .
serious blunder on my part, I don't know what I was thinking :/
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