mersenneforum.org Arithmetic progressions for n
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 2017-05-05, 18:55 #1 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 2·7·43 Posts Arithmetic progressions for n HI Mersenneforum, Some are familiar with the technique of arithmetic progression. We are familiar with the sum of an arithmetic progression For example - 1+2+3+ +n ~ n*(n+1)/2 similarly 4+5+ +n ~ (n+4)*(n+5)/2. Now I am relatively certain that the line above is true. Okay I did not tap the equals sign. The technique of arithmetic progression goes like this. Assume the Base case is true. Write it for n is 0. Then write that the n'th case implies the (n+1)'st case. End lecture. For what it's worth. Regards, Matt
2017-05-05, 19:17   #2
science_man_88

"Forget I exist"
Jul 2009
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Quote:
 Originally Posted by MattcAnderson HI Mersenneforum, Some are familiar with the technique of arithmetic progression. We are familiar with the sum of an arithmetic progression For example - 1+2+3+ +n ~ n*(n+1)/2 similarly 4+5+ +n ~ (n+4)*(n+5)/2. Now I am relatively certain that the line above is true. Okay I did not tap the equals sign. The technique of arithmetic progression goes like this. Assume the Base case is true. Write it for n is 0. Then write that the n'th case implies the (n+1)'st case. End lecture. For what it's worth. Regards, Matt
you seems to be talking closer to a proof technique called mathematical induction. although you can mathematically induce the formulae in theory admittedly.

4+5+ +n is 6 less than 1+2+3+ +n ~ n*(n+1)/2 so the logic could go 4+5+ +n ~(n*(n+1)-12)/2) = (n^2+n-12) /2 = (used PARI/GP) ((n+4)*(n-3))/2 edit: if you want to talk in general you can do y+(y+1)+...+n by realizing it's ((y-1)*y)/2 less than the value you want to put that over two you multiply it by 2 ( so it's net value doesn't change you get (y-1)*y subtract that from (n*(n+1)) and get (n*(n+1)-(y-1)*y)/2 then factor the numerator (n^2+n-(y-1)*y) into (n+y)*(n-(y-1)) = n^2+(y*n-(y-1)*n)-(y-1)*y undistribute the n in the parentheses and get n^2+n-(y-1)*y just like we thought so it's (n+y)*(n*(y-1))/2 where y is the number you start with consecutively. edit: this falls under univariate linear equations on khan academy and though 1+2+3+4+... ( equally spaced) is a sum of an arithmetic progression so is 1+7+13+19+.... ( as before equally spaced) it just so happens the sum of an arithmetic progression is the sum of the first and last number times the number of pairs ( aka the number of terms being summed divided by 2) so the formula works as (n+1)*n/2 because 1 is the first term and there's n terms so there are n/2 pairings. so again for your 4 example you get start number =4; sum of first and last = n+4; number of terms =n-3; so you get (n+4)*(n-3)/2

Last fiddled with by science_man_88 on 2017-05-05 at 20:07

2017-05-06, 08:54   #3
LaurV
Romulan Interpreter

Jun 2011
Thailand

212478 Posts

Quote:
 Originally Posted by MattcAnderson similarly 4+5+ +n ~ (n+4)*(n+5)/2. Now I am relatively certain that the line above is true.
nope...

 2017-05-06, 12:22 #4 science_man_88     "Forget I exist" Jul 2009 Dumbassville 8,369 Posts oh and for further clarification on the name they are called arithmetic series.
 2017-05-06, 14:10 #5 Dr Sardonicus     Feb 2017 Nowhere 52·11·13 Posts Here is a curiosity about APs of consecutive integers. EXERCISE: Prove that an integer greater than 1 is the sum of two or more consecutive positive integers, if and only if it is not a power of 2. Last fiddled with by Dr Sardonicus on 2017-05-06 at 14:11
2017-05-06, 14:34   #6
science_man_88

"Forget I exist"
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Quote:
 Originally Posted by Dr Sardonicus Here is a curiosity about APs of consecutive integers. EXERCISE: Prove that an integer greater than 1 is the sum of two or more consecutive positive integers, if and only if it is not a power of 2.
like in the restatement of Goldbach's conjecture to the form that every natural number above a certain point must be equidistant to two primes, I believe the proof would be something like the following:

let a=2*b be a natural number then any two numbers that add to a must be equidistant from b
but for two consecutive natural numbers to be equidistant from a value of b it must a half integer aka a mixed number that ends with 1/2 which only happens if a is odd or one of its divisors is ( aka not a pure even integer aka not a power of two) in the case of three numbers adding to a number they can add to three times the middle number but that 3 defeats the ability to be a pure even number similar arguments can disprove an odd number of summands. and any number divisible by an odd number can be split in such a manner the only numbers that don't divide by an odd number greater than 1 are the powers of two so they are the only numbers not to be a sum of consecutive numbers.
am I even close. edit:doh much simpler:

(n+1)*n/2 only one of n+1 or n is even so there will still be an odd factor.
meaning it can't sum to a pure even number sorry head is wacko again I guess.

Last fiddled with by science_man_88 on 2017-05-06 at 14:54

 2017-05-06, 19:08 #7 science_man_88     "Forget I exist" Jul 2009 Dumbassville 8,369 Posts yes I know I still fumbled the proof a bit for the general case. Arithmetic progressions can be combined with geometric progressions to form arithmetico-geometric sequences etc. Last fiddled with by science_man_88 on 2017-05-06 at 19:34
 2017-05-06, 21:57 #8 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 192910 Posts It's a good thing. Any odd number o' (including 1): o' = o'\2+(o'\2+1) ----- Any even number e, not a power of 2 will be divisible by an odd number o. Then e= (e/o-o\2)+...+e/o+(e/o+1)+...+(e/o+o\2) ----- For t=2^x if there exists a sequence of consecutive integers n to m inclusive that sum up to t, then t = (n+m)(m-n+1)/2 if (n+m) is odd then (m-n+1) is even if (n+m) is even then (m-n+1) is odd That means that t is either the product of an even and an odd integer or, t is the product of an odd and an odd integer Since no power of 2 can be the product of an odd integer, then there can be no summation of consecutive integers that equals any powers of 2. Last fiddled with by a1call on 2017-05-06 at 22:37
2017-05-07, 00:06   #9
a1call

"Rashid Naimi"
Oct 2015
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3·643 Posts

Missing:

For integer x > 0
Corrected version:

Quote:
 Originally Posted by a1call It's a good thing. Any odd integer o' (including 1): o' = o'\2+(o'\2+1) ----- Any even integer e, not a power of 2 will be divisible by an odd integer o. Then e= (e/o-o\2)+...+e/o+(e/o+1)+...+(e/o+o\2) ----- For t=2^x where x is an integer greater than 0 if there exists a sequence of consecutive integers n to m inclusive that sum up to t, then t = (n+m)(m-n+1)/2 if (n+m) is odd then (m-n+1) is even if (n+m) is even then (m-n+1) is odd That means that t is either the product of an even and an odd integer or, t is the product of an odd and an odd integer Since no power of 2, greater than 1 can be the product of an odd integer, then there can be no summation of consecutive integers that equals any powers of 2, greater than 1.

Last fiddled with by a1call on 2017-05-07 at 00:51

 2017-05-07, 01:17 #10 science_man_88     "Forget I exist" Jul 2009 Dumbassville 8,369 Posts for other properties see: https://en.wikipedia.org/wiki/Arithm...rogression#Sum
 2017-05-07, 01:58 #11 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 192910 Posts It's missing the historical section and I can't find it anywhere. I recall a tale told by one of my math teachers as to how a mathematician as a young pupil summed up numbers from 1 to 100 very quickly. A task which took his classmates much much longer.

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