THats why i told u can u explain it to me in another way . Its just that what u have told in POST#12 reagrding my n-1 doubt i am trying since then to make a self explantion statment.
Arithmetic progression
- Thread starter Saumyojit
- Start date
@Dr.Peterson My equation stands like this
550/2=n/2(20+30)
n=11
why do u say "you are taking n to mean only half the number of one-way trips" -->11 means 11 one way trips.
@Dr.Peterson @JeffM First see the video between 3:37 sec to 4:48 sec carefully
Eddie is deriving the formula S=n/2(first term +Last term) . He took eg of this ap= 1+3+5+7+9+11
He created blocks of each term. Then for the sake of derivation he duplicated one extra block for every each block/term of the orignal series.
(i)Now at 3:37 he mixed the first block (first term of the 1st series) with the last block (last term of the 2nd series). Why Did not he mixed the first block of the 1st series with any other block of the 2nd series except the last one ;That's because he wanted to get the sum of every combination same or the height of every block same . So mixing first with last,2nd with 2nd Last gives me a equal sum (12) in every case . Right?
Main doubt: (ii)At 4:08 He says why mixing is more helpful he says "Becoz the series is in Ap (How it grows is the same as how it shrinks)"---- I Understand that every term grows by a cd or term shrinks by cd but still I cannot understand his quote "Becoz the series is in Ap (How it grows is the same as how it shrinks)" What it has to do with mixing of 2 blocks .
I understand the other reason . we have the Formula n(first term +Last term) in R.H.S before divide by 2 operation so when we mix 2 blocks we get the same height of every block with which i can genralize all the other blocks to ONE BLOCK only and can multiply it with no of blocks present . It makes a lot easier to find the TOTAL Size of the rectangular array.
He says that at 4:48 there are 6 columns of the same height so this pair(1+11) can be multiplied by 6 to be equated with total sum(back & forth).
Eddie is deriving the formula S=n/2(first term +Last term) . He took eg of this ap= 1+3+5+7+9+11
He created blocks of each term. Then for the sake of derivation he duplicated one extra block for every each block/term of the orignal series.
(i)Now at 3:37 he mixed the first block (first term of the 1st series) with the last block (last term of the 2nd series). Why Did not he mixed the first block of the 1st series with any other block of the 2nd series except the last one ;That's because he wanted to get the sum of every combination same or the height of every block same . So mixing first with last,2nd with 2nd Last gives me a equal sum (12) in every case . Right?
Main doubt: (ii)At 4:08 He says why mixing is more helpful he says "Becoz the series is in Ap (How it grows is the same as how it shrinks)"---- I Understand that every term grows by a cd or term shrinks by cd but still I cannot understand his quote "Becoz the series is in Ap (How it grows is the same as how it shrinks)" What it has to do with mixing of 2 blocks .
I understand the other reason . we have the Formula n(first term +Last term) in R.H.S before divide by 2 operation so when we mix 2 blocks we get the same height of every block with which i can genralize all the other blocks to ONE BLOCK only and can multiply it with no of blocks present . It makes a lot easier to find the TOTAL Size of the rectangular array.
He says that at 4:48 there are 6 columns of the same height so this pair(1+11) can be multiplied by 6 to be equated with total sum(back & forth).
You keep trying to understand numbers by understanding other things, a process that detracts from thinking about numbers as entities in their own right. The video is a very clever visual, but it is about numbers. You have to take the visual display and turn it into numbers to develop an understanding. The video flies by and makes things look obvious. You still have to work it out step by step in terms of numbers.
x = 1 + 3 + 5 + 7 + 9 + 11.
He doubles everything
2x =
1 + 3 + 5 + 7 + 9 + 11 +
1 + 3 + 5 + 7 + 9 + 11.
He does his reversing visual to get
2x =
1 + 3 + 5 + 7 + 9 + 11 +
11 + 9 + 7 + 5 + 3 + 1.
Start adding those two rows up by adding them down. You get:
(1 + 11) + (3 + 9) + (5 + 7) + (7 + 5) + (9 + 3) + (11 + 1).
Now what he flies by is this. With respect to any successive pair of parentheses, he notes, almost in passing, that
(a + b) + (c + d) = (a + b) + {(a + 2) + (b - 2)} =
(a + b) + (a + b) + (2 - 2) = 2(a + b).
This is true because, when he did the reversing, the terms are increasing in the first row and decreasing in the second row by a constant term. When we add down all those constant terms simply become a bunch of zeroes. Thus, each parenthetical term is equal to the preceding one, and we can replace them all with the first one.
2x =
(1 + 11) + (1 + 11) + (1 + 11) + (1 + 11) + (1 + 11) + (1 + 11) =
6(1 + 11).
Now he divides by 2 to get back to x.
[MATH]x = \dfrac{6}{2} * (1 + 12).[/MATH]
Now he generalizes to the sum of an arithmetic progression is equal to
half the number of terms times the sum of the first and last terms.
This IS NOT A PROOF. It is a very clever, very quick way to make a formula plausible. If you want to understand it, you must figure out what it means in terms of the underlying numbers. If you want to prove it, you must use algebra to prove it for any arithmetic progression starting at a with n terms and a constant difference of d.
x = 1 + 3 + 5 + 7 + 9 + 11.
He doubles everything
2x =
1 + 3 + 5 + 7 + 9 + 11 +
1 + 3 + 5 + 7 + 9 + 11.
He does his reversing visual to get
2x =
1 + 3 + 5 + 7 + 9 + 11 +
11 + 9 + 7 + 5 + 3 + 1.
Start adding those two rows up by adding them down. You get:
(1 + 11) + (3 + 9) + (5 + 7) + (7 + 5) + (9 + 3) + (11 + 1).
Now what he flies by is this. With respect to any successive pair of parentheses, he notes, almost in passing, that
(a + b) + (c + d) = (a + b) + {(a + 2) + (b - 2)} =
(a + b) + (a + b) + (2 - 2) = 2(a + b).
This is true because, when he did the reversing, the terms are increasing in the first row and decreasing in the second row by a constant term. When we add down all those constant terms simply become a bunch of zeroes. Thus, each parenthetical term is equal to the preceding one, and we can replace them all with the first one.
2x =
(1 + 11) + (1 + 11) + (1 + 11) + (1 + 11) + (1 + 11) + (1 + 11) =
6(1 + 11).
Now he divides by 2 to get back to x.
[MATH]x = \dfrac{6}{2} * (1 + 12).[/MATH]
Now he generalizes to the sum of an arithmetic progression is equal to
half the number of terms times the sum of the first and last terms.
This IS NOT A PROOF. It is a very clever, very quick way to make a formula plausible. If you want to understand it, you must figure out what it means in terms of the underlying numbers. If you want to prove it, you must use algebra to prove it for any arithmetic progression starting at a with n terms and a constant difference of d.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,725
The key idea is that when the sequence is reversed, one copy increases by the same amount the other increases, so that the sum is constant.
What JeffM just wrote is what I would have said next, so I'll leave you read that. He has some very wise comments.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,725
Correction (if it wasn't obvious):
The key idea is that when the sequence is reversed, one copy increases by the same amount the other decreases, so that the sum is constant.
I am going to give a formal proof that depends on this theorem, which I will not prove.
[MATH]1 + 2 +\ ... \ m = \dfrac{m(m + 1)}{2}.[/MATH]
If you doubt that, pick a dozen values for m and convince yourself. (A true proof probably is a bit too advanced for you.)
[MATH]x = (a + 0 * d) + (a + d) + (a + 2d) + \ ... \ \{a + (n - 1)d\}.[/MATH]
In other words, the first term is
[MATH]\alpha = \alpha + 0 * d.[/MATH]
The last term is
[MATH]\omega = \alpha + (n - 1)d.[/MATH]
That is one way to understand where the the (n - 1) that you asked about before comes from. Notice we have n terms, but the final term differs from the latter by (n - 1)d, where d is the common difference between successive terms of the sequence.
[MATH]\therefore x = an + \{(0 * d) + (1 * d) + 2d \ ... \ (n - 1)d\} \implies[/MATH]
[MATH]x = an + \{1d + 2d + \ ... \ (n - 1)d\} \implies[/MATH]
[MATH]x = an + d(1 + 2 + \ ... \ n - 1) \implies[/MATH]
[MATH]x = an + d * \dfrac{(n - 1)\{(n - 1) + 1\}}{2} \implies[/MATH]
[MATH]x = n * \left (a + \dfrac{(n - 1)d}{2} \right ) = \dfrac{n}{2} * \{ 2a + (n - 1)d\} \implies[/MATH]
[MATH]x = \dfrac{n}{2} * (a + \{ a + (n - 1)d\}) = \dfrac{n}{2} * ( \alpha + \omega)[/MATH]
That is a formal proof depending on one theorem and basic properties of numbers. No analogies. No linguistic issues. Just logic.
[MATH]1 + 2 +\ ... \ m = \dfrac{m(m + 1)}{2}.[/MATH]
If you doubt that, pick a dozen values for m and convince yourself. (A true proof probably is a bit too advanced for you.)
[MATH]x = (a + 0 * d) + (a + d) + (a + 2d) + \ ... \ \{a + (n - 1)d\}.[/MATH]
In other words, the first term is
[MATH]\alpha = \alpha + 0 * d.[/MATH]
The last term is
[MATH]\omega = \alpha + (n - 1)d.[/MATH]
That is one way to understand where the the (n - 1) that you asked about before comes from. Notice we have n terms, but the final term differs from the latter by (n - 1)d, where d is the common difference between successive terms of the sequence.
[MATH]\therefore x = an + \{(0 * d) + (1 * d) + 2d \ ... \ (n - 1)d\} \implies[/MATH]
[MATH]x = an + \{1d + 2d + \ ... \ (n - 1)d\} \implies[/MATH]
[MATH]x = an + d(1 + 2 + \ ... \ n - 1) \implies[/MATH]
[MATH]x = an + d * \dfrac{(n - 1)\{(n - 1) + 1\}}{2} \implies[/MATH]
[MATH]x = n * \left (a + \dfrac{(n - 1)d}{2} \right ) = \dfrac{n}{2} * \{ 2a + (n - 1)d\} \implies[/MATH]
[MATH]x = \dfrac{n}{2} * (a + \{ a + (n - 1)d\}) = \dfrac{n}{2} * ( \alpha + \omega)[/MATH]
That is a formal proof depending on one theorem and basic properties of numbers. No analogies. No linguistic issues. Just logic.
Last edited by a moderator:
D
Deleted member 4993
Guest
The proof is simple and elegant (as follows)!
S = 1...+...2...+...3...+ ........+...(m-2) + (m-1) + m .............................(1) flip the sequence and re-write
S = m+(m-1)+(m-2)+.....+ .......3.....+ ....2......+ 1 ..............................(2) .....................Add (1) & (2) paired vertically
________________________________________________
2S = (1+m) + (1+m) + (1+m).......(1+m) + (1+m) + (1+m)
2S = m * (1 + m)
S = (m/2) * (1+m)
I believe Cauchy (or Gauss or Euler) added 1 through 100 by this technique in 2nd grade.
This and similar "simple" tricks come very handy in this journey.....