anthony@utcsstat.uucp (Anthony Ayiomamitis) (01/23/86)
Can anyone deduce the sum of the following series after n terms?
1 + 3 + 6 + 10 + 15 + 21 + .....
--
{allegra,ihnp4,linus,decvax}!utzoo!utcsstat!anthony
{ihnp4|decvax|utzoo|utcsrgv}!utcs!utzoo!utcsstat!anthonybs@linus.UUCP (Robert D. Silverman) (01/24/86)
> > Can anyone deduce the sum of the following series after n terms? > > 1 + 3 + 6 + 10 + 15 + 21 + ..... > > -- > > {allegra,ihnp4,linus,decvax}!utzoo!utcsstat!anthony > {ihnp4|decvax|utzoo|utcsrgv}!utcs!utzoo!utcsstat!anthony The n'th term is: n(n+1)(n+2) ----------- 6 Bob Silverman
ark@alice.UucP (Andrew Koenig) (01/24/86)
Sure ... the sum of n terms of 1 + 3 + 6 + 10 + 15 ... is n(n+1)(n+2)/6
hes@ecsvax.UUCP (01/24/86)
> > Can anyone deduce the sum of the following series after n terms? > > 1 + 3 + 6 + 10 + 15 + 21 + ..... > -- > {allegra,ihnp4,linus,decvax}!utzoo!utcsstat!anthony This appears to be (using S for capital sigma): n k S S i k=1 i=1 The second summation = k(k+1)/2, and then the first summation can be performed by multiplying this out, separating the terms, using this same relationship again plus this one: n S k^2 = n(n+1)(2n+1)/6 k=1 --henry schaffer
pnp@ihnp1.UUCP (Pete Prokopowicz) (01/24/86)
1 + 3 + 6 + 10 + ... + = ? the sum of n terms is (n**2 + n) x (n + 2) / 6 I think.
percus@acf4.UUCP (Allon G. Percus) (01/24/86)
> Can anyone deduce the sum of the following series after n terms? > > 1 + 3 + 6 + 10 + 15 + 21 + ..... If I am correct, n (n + 1) (n + 2) S = ----------------- 6 should simply work. . ------- |-----| A. G. Percus |II II| (ARPA) percus@acf4 |II II| (NYU) percus.acf4 |II II| (UUCP) ...{allegra!ihnp4!seismo}!cmcl2!acf4!percus |II II| -------
weemba@brahms.BERKELEY.EDU (Matthew P. Wiener) (01/24/86)
> > Can anyone deduce the sum of the following series after n terms? > > 1 + 3 + 6 + 10 + 15 + 21 + ..... > yes ucbvax!brahms!weemba Matthew P Wiener/UCB Math Dept/Berkeley CA 94720
gwyn@brl-tgr.UUCP (01/26/86)
> > Can anyone deduce the sum of the following series after n terms? > > > > 1 + 3 + 6 + 10 + 15 + 21 + ..... > > yes Well, almost anyone. Actually, one has to know how to look things up in standard reference books, or else be able to do simple math.
stassen@spp2.UUCP (Chris Stassen) (01/26/86)
In article <2265@utcsstat.uucp> anthony writes: > > Can anyone deduce the sum of the following series after n terms? > > 1 + 3 + 6 + 10 + 15 + 21 + ..... S(n) = n ( n^2 + 3n + 2 ) / 6 I have a simple method that works for all series that can be expressed as polymonials. I developed it on my own in High School, but I am sure that it is an accepted method (and that someone else has named it). In fact, there is probably a less clumsy version of the method, but I haven't read anything about it. (Any pointers to books would be accepted - I like this stuff). 1 3 6 10 15 21 28 36 45 55 - Set of terms 1 4 10 20 35 56 - First few sums 3 6 10 15 21 - Difference between adjacent sums (n^1) 3 4 5 6 - Difference between adjacent diffs (n^2) 1 1 1 - Difference between adjacent diffs (n^3) When we get constants across a row, then that row is the highest power of n involved in the expression for the sum (all rows that follow would be zero). The factor of n^m (in constant row "m") is C/(m!), where C is the constant (the value of all numbers in that row). Therefore, n^3/6 is the first part of the polynomial. Now, we have to subtract n^3/6 from each sum. n^3/6 5/6 16/6 33/6 56/6 85/6 120/6 - First few sums (minus n^3/6) 11/6 17/6 23/6 29/6 35/6 - Difference btw adj sums (n) 6/6 6/6 6/6 6/6 - Difference btw adj diff (n^2) Note that the n^2 row is now the constant row (we have subtracted out the n^3 part of the polynomial expression). C=1, and the coefficient for n^2 is therefore 1/2. Now, we have to subtract n^2/2 from each sum. n^2/2 2/6 4/6 6/6 8/6 10/6 - First few sums (less n^3/6 and n^2/2) 2/6 2/6 2/6 2/6 - Difference between adjacent sums (n) Note that the n row is now the constant row (we have subtracted out the n^3 and n^2 parts of the polynomial expression). C=1/3, and the coefficient of n is therefore 1/3. Now, we subtract n/3 from each sum. n/3 0 0 0 0 0 - Sums There is no constant in the polynomial. Therefore, the expression for the sum is: n ( n^2 + 3n + 2 ) / 6 Let's test it to be sure. 1 4 10 20 35 56 - First few sums n = 1 : S(n) = 1 ( 1 + 3 + 2 ) / 6 = 1 n = 2 : S(n) = 2 ( 4 + 6 + 2 ) / 6 = 4 n = 3 : S(n) = 3 ( 9 + 9 + 2 ) / 6 = 10 n = 4 : S(n) = 4 (16 +12 + 2 ) / 6 = 20 -- Chris
eirik@tekchips.UUCP (Eirik Fuller) (01/27/86)
In article <2265@utcsstat.uucp> anthony@utcsstat.uucp (Anthony Ayiomamitis) writes: > > Can anyone deduce the sum of the following series after n terms? > > 1 + 3 + 6 + 10 + 15 + 21 + ..... > >-- > > {allegra,ihnp4,linus,decvax}!utzoo!utcsstat!anthony > {ihnp4|decvax|utzoo|utcsrgv}!utcs!utzoo!utcsstat!anthony Well, that depends on what the rest of the terms are ... 8^). The Nth partial sum of this series is S(N) = N(N+1)(N+2)/6. Proof: True for N=1 (check it out). S(N) - S(N-1) = N(N+1)[(N+2)-(N-1)]/6 = N(N+1)/2, the Nth term in the sum. By induction, the formula holds. Oh yeah, to convince yourself that the Nth term is N(N+1)/2 : The most obvious pattern I detect in the terms is that the Nth term differs from the (N-1)th term by N, so the Nth term is the sum of the first N integers. Using the same idea as above, N=1 checks, and [N(N+1)/2] - [(N-1)N/2] = N[(N+1)-(N-1)]/2 = N. Sorry for the length of this (well, maybe ... :-), but I have two more observations. Long ago I worked out a formula for summing the first N integers, each raised to the Kth power (sum from I = 1 to N of [I to the K], in clumsy ASCII notation :-), where K is a positive integer. It involves solving a linear set of equations with Pascal's triangle in a triangular matrix; beautiful tidbit, or so I thought when I discovered it. With even the slightest bit of encouragement (DON'T POST; mail only please) I could be pursuaded to post it, but I fear it might be common knowledge. Anyway, this formula is what I used to find the answer to this problem (I had to re-derive it though). The other observation. My crack about knowing the other terms in the series was half serious. (Seriesly ... :-). Given any series specified by a finite number of terms, it is trivially easy to generate a closed form equation for an infinite number of other series beginning with the same finite set of initial terms. So which of all possible series is meant in any one case? It depends on being able to recognize the pattern. Of course, I am presumptuous enough to assume that I guessed the intent of this one. My formula works for all of the given terms anyway 8^). One more brief (?) closing comment: my original excursion into this was triggered by elementary integration in beginner calculus. These formulas can be used (without the fundamental theorem of calculus) to integrate polynomials using the definition of integration as a limiting sum of rectangles. Any calculus text I've seen has these formulas for K=1,2, and 3, generally with little or no proof.
stark@sbcs.UUCP (Eugene Stark) (01/27/86)
> > Can anyone deduce the sum of the following series after n terms? > > 1 + 3 + 6 + 10 + 15 + 21 + ..... > > -- > > {allegra,ihnp4,linus,decvax}!utzoo!utcsstat!anthony > {ihnp4|decvax|utzoo|utcsrgv}!utcs!utzoo!utcsstat!anthony Assuming that you want the sum of the first n triangular numbers: n i -- -- \ \ / / j = n(n+1)(n+2)/6 -- -- i=1 j=1 You can derive this yourself if you use the fact that the sum n -- \ k / i -- i=1 is a polynomial in n of order k+1. Or, you can look up in a book the sums for k = 1 and k = 2, namely n(n+1)/2 and n(n+1)(2n+1)/6. Gene Stark SUNY at Stony Brook
reichel@newton.DEC (01/28/86)
--------------------Reply to mail dated 24-JAN-1986 19:46--------------------
stephen@datacube.UUCP (01/28/86)
> Can anyone deduce the sum of the following series after n terms? > > 1 + 3 + 6 + 10 + 15 + 21 + ..... Yeah, its: n*(n^2 + 3*n + 2)/6. Stephen Watkins UUCP: ihnp4!datacube!stephen Datacube Inc.; 4 Dearborn Rd.; Peabody, Ma. 01960; 617-535-6644
sher@rochester.UUCP (David Sher) (01/29/86)
Regarding the sum of the sums of the first k integers k goes from 1 to n: How many out there solved it by considerring the volume contained in a maximal corner of a cube of size n (the largest volume that you can get by cutting off a corner of a cube without removing any other corners). A little thought will make the relationship between the mathematical and geometric entities obvious. (Of course this is high school math (I was doing this kind of stuff in high school anyway)). -- -David Sher sher@rochester seismo!rochester!sher
weemba@brahms.BERKELEY.EDU (Matthew P. Wiener) (01/30/86)
>Regarding the sum of the sums of the first k integers k goes from 1 to n: >How many out there solved it by considerring the volume contained in a >maximal corner of a cube of size n (the largest volume that you can get by >cutting off a corner of a cube without removing any other corners). >A little thought will make the relationship between the mathematical >and geometric entities obvious. (Of course this is high school math >(I was doing this kind of stuff in high school anyway)). And how many solved it by: inverse finite differences? Newton interpolation? finding a cubic spline? undetermined coefficients? using Pade approximants? identifying the generating function? Euler-Maclaurin summation? running symbolic algebra software? looking it up in a book? recalling the answer instantly? Who said the problem had to be easy? ucbvax!brahms!weemba Matthew P Wiener/UCB Math Dept/Berkeley CA 94720
mouse@mcgill-vision.UUCP (der Mouse) (01/30/86)
>> Can anyone deduce the sum of the following series after n terms? >> 1 + 3 + 6 + 10 + 15 + 21 + ..... > I have a simple method that works for all series that can > be expressed as polymonials. I developed it on my own in High > School, but I am sure that it is an accepted method (and that > someone else has named it). (Any pointers to books would be > accepted - I like this stuff). > 1 4 10 20 35 56 - First few sums > 3 6 10 15 21 - Difference between adjacent sums (n^1) > 3 4 5 6 - Difference between adjacent diffs (n^2) > 1 1 1 - Difference between adjacent diffs (n^3) Ah, but would pointers to books be welcomed? (:-) I saw this in one of Martin Gardner's books, probably one of his [N, N=1, 2, ...] th Book of Scientific American Mathematical Puzzles and Diversions, or some such title. You can probably find it by looking him up as an author (Books In Print, or the Author section of the card catalog). It was called the Calculus of Finite Differences, and an anecdote was told: "When I was [n, n~=10 to 15] years old, I noticed that if you took the differences between the squares of the numbers, and then again, you got a constant. I showed this to my father, and in his good Quaker English, he said, 'Why John, thee hast discovered the calculus of finite differences.'" -- I forget who the speaker is in this quote. I did something similar: I realized that the C of F D guaranteed that a cubic would be sufficient, but not remembering the formula and not feeling like deriving it, I wrote out a + b + c + d = 1 [ from an^3+bn^2+cn+d = S, n=1 ] 8a + 4b + 2c + d = 4 [ ... and n=2 ... ] 27a + 9b + 3c + d = 10 [ n=3 ] 64a + 16b + 4c + d = 20 [ n=4 ] and solved.... -- der Mouse USA: {ihnp4,decvax,akgua,etc}!utcsri!mcgill-vision!mouse philabs!micomvax!musocs!mcgill-vision!mouse Europe: mcvax!decvax!utcsri!mcgill-vision!mouse mcvax!seismo!cmcl2!philabs!micomvax!musocs!mcgill-vision!mouse Hacker: One who accidentally destroys / Wizard: One who recovers it afterward
cralle@lll-crg.ARpA (Bob Cralle) (02/03/86)
> > I did something similar: I realized that the C of F D guaranteed >that a cubic would be sufficient, but not remembering the formula and >not feeling like deriving it, I wrote out > > a + b + c + d = 1 [ from an^3+bn^2+cn+d = S, n=1 ] > 8a + 4b + 2c + d = 4 [ ... and n=2 ... ] >27a + 9b + 3c + d = 10 [ n=3 ] >64a + 16b + 4c + d = 20 [ n=4 ] > What is the next term in the series: 1 2 4 8 16 ... ? If I told you I wanted 31 what would you say? Since, as has been pointed out, I can make it what ever I want, what do I get by your method? (Thirty years ago when I solved this, I justed differenced it.) At the end you may be interested to see what the series comes from. As math types need MACSYMA more than they sometimes admit I let it do all of the algebra. (c3) a*n^5+b*n^4+c*n^3+d*n^2+e*n+f=g; 5 4 3 2 (d3) a n + b n + c n + d n + e n + f = g (c4) d3,n:1,g:1; (d4) f + e + d + c + b + a = 1 (c5) d3,n:2,g:2; (d5) f + 2 e + 4 d + 8 c + 16 b + 32 a = 2 (c6) d3,n:3,g:4; (d6) f + 3 e + 9 d + 27 c + 81 b + 243 a = 4 (c7) d3,n:4,g:8; (d7) f + 4 e + 16 d + 64 c + 256 b + 1024 a = 8 (c8) d3,n:5,g:16; (d8) f + 5 e + 25 d + 125 c + 625 b + 3125 a = 16 (c9) d3,n:6,g:31; (d9) f + 6 e + 36 d + 216 c + 1296 b + 7776 a = 31 (c10) solve([d4,d5,d6,d7,d8,d9]); 3 23 1 1 (d10) [[f = 1, e = - -, d = --, c = - -, b = --, a = 0]] 4 24 4 24 (c13) d10[1]; 3 23 1 1 (d13) [f = 1, e = - -, d = --, c = - -, b = --, a = 0] 4 24 4 24 (c19) subst(":","=",d13); 3 23 1 1 (d19) [f : 1, e : - -, d : --, c : - -, b : --, a : 0] 4 24 4 24 (c22) d3,d19; 4 3 2 n n 23 n 3 n (d22) -- - -- + ----- - --- + 1 = g 24 4 24 4 (c28) factor(lhs(d22)); 4 3 2 n - 6 n + 23 n - 18 n + 24 (d29) ----------------------------- 24 and so the series goes: 1 2 4 8 16 31 57 99 163 256 386 562... [isn't the 256 curious--one late!] What does the series represent? If you have n points on a circle & and you connect every point to every other point the series gives the number of areas those lines cut the circle into. (more than two lines intersecting at a point not allowed.) Anyone desirous of a copy of a paper I wrote on this subject & a proof, send me your address. -- Regards, &c., Bob cralle@lll-crg R. K. Cralle LLNL MS L-73 POX 808 Livermore, CA 94550 fone 415/422-4041 FTS 415/532-4041 <<< Facts are no match for belief >>>