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Algebra
 Summation of Series by the Method of Differences
  Example
   Example
 Direct Factorial Series
  Example
  To find the sum of 𝑛 terms
 Inverse Factorial Series
  Example
  Example
  Example
 Composite Factorial Series
 Miscellaneous Series
  Summation of a series partly Arithmetical and partly Geometrical
   Example
 Sources and References

Algebra

Summation of Series by the Method of Differences

Rule: From successive series of differences until a series of equal differences is obtained. Let 𝑎, 𝑏, 𝑐, 𝑑, ⋯ be the first terms of the several series; 264 then the 𝑛th term of the given series is 𝑎+(𝑛−1)𝑏+(𝑛−1)(𝑛−2)1⋅2𝑐+(𝑛−1)(𝑛−2)(𝑛−3)1⋅2⋅3𝑑+265 The sum of 𝑛 terms =𝑛𝑎+𝑛(𝑛−1)1⋅2𝑏+𝑛(𝑛−1)(𝑛−2)1⋅2⋅3𝑐+⋯ Proved by Induction266

Example

𝑎1+5+15+35+70+126+ 𝑏4+10+20+35+56+ 𝑐6+10+15+21+ 𝑑4+5+6+ 𝑒1+1+ The 100th term of the first series =1+99⋅4+99⋅981⋅26+99⋅98⋅971⋅2⋅34+99⋅98⋅97⋅961⋅2⋅3⋅4 The sum of 100 terms =100+100⋅991⋅24+100⋅99⋅981⋅2⋅36+100⋅99⋅98⋅971⋅2⋅3⋅44+100⋅99⋅98⋅97⋅961⋅2⋅3⋅4⋅5266 To interpolate a term between two terms of a series by the method of differences.

Example

Given log 71, log 72, log 73, log 74, it is required to find log 72⋅54. Form the series of differences from the given logarithms, as in (266).  log 71log 72log 73log 74 𝑎⋯1.85125831.85733251.86332291.8692317 𝑏⋯.0060742.0059904.0059088 𝑐⋯−.0000838−.0000816 𝑑⋯−.000022 considered to vanish. log 72⋅54 must be regarded as an interpolated term, the number of its place being 2⋅54. Therefore put 2⋅54 for 𝑛 in formula (265). Result log 72⋅54=1⋅8605777 267

Direct Factorial Series

Example

Ex 5⋅7⋅9+7⋅9⋅11+9⋅11⋅13+11⋅13⋅15+⋯ 𝑑=common difference of factors 𝑚=number of factors in each term 𝑛=number of terms 𝑎=first factor of first term −𝑑 𝑛th term=(𝑎+𝑛𝑑)(𝑎+𝑛+1𝑑)⋯(𝑎+𝑛+𝑚−1𝑑)268

To find the sum of 𝑛 terms

Rule: From the last term with the next highest factor take the first term with the next lowest factor, and divide by (𝑚+1)𝑑. Proof by Induction. Thus the sum of 4 terms of the above series will be, putting 𝑑=2, 𝑚==3, 𝑛=4, 𝑎==3, 𝑆=11⋅13⋅15⋅17−3⋅5⋅7⋅9(3+1)2 Proved either by Induction, or by the method of Indeterminate Coefficients. 269

Inverse Factorial Series

Example

Ex. 15⋅7⋅9+17⋅9⋅11+19⋅11⋅13+111⋅13⋅15+⋯ Defining 𝑑, 𝑚, 𝑛, 𝑎 as before, the 𝑛th term=1(𝑎+𝑛𝑑)(𝑎+𝑛+1𝑑)⋯(𝑎+𝑛+𝑚−1𝑑) 270 To find the sum of 𝑛 terms. Rule.: From the first term wanting its last factor take the last term wanting its first factor, and divide by (𝑚−1)𝑑. Thus the sum of 4 terms of the above series will be, putting 𝑑=2, 𝑚=3, 𝑛=4, 𝑎=3, 15⋅7113⋅15(3−1)2 Proof: By Induction, or by decomposing theterms, as in the following example. 271

Example

Ex. To sum the same series by decomposing the terms into partial fractions. Take the general term in the simple form 2(𝑟−2)𝑟(𝑟+2) Resolve this into the three fractions 18(𝑟−2)14𝑟+18(𝑟+2) by (235) Substitute 7, 9, 11, ⋯ successively for 𝑟, and the given series has for its equivalent the three series 18{ 15+17+19+111+113++12𝑛+3} +18{272921121312𝑛+312𝑛+5} +18{ 19+111+113++12𝑛+3+12𝑛+5+12𝑛+7} and the sum of 𝑛 terms is seen, by inspection, to be 18{151712𝑛+5+12𝑛+7}=14{15⋅71(2𝑛+5)(2𝑛+7)} a result obtained at once by the rule in (271), taking 15⋅7⋅9 for the first term, and 1(2𝑛+3)(2𝑛+5)(2𝑛+7) for the 𝑛th or last term.272 Analogous series may be reduced to the types in (268) and (270), or else the terms may be decomposed in the manner shewn in (272).

Example

Ex: 11⋅2⋅3+42⋅3⋅4+73⋅4⋅5+104⋅5⋅6+⋯ has for its general term 3𝑛−2𝑛(𝑛+1)(𝑛+2)=−1𝑛+5𝑛+14𝑛+2 by (235) and we may proceed as in (272) to find the sum of 𝑛 terms. The method of (272) includes the method known as "Summation by Subtraction." The method of (272) includes the method known as "Summation by Subtraction", but it has the advantage of being more general and easier of application to complex series. 273

Composite Factorial Series

If the two series (1−𝑥)−5=1+5𝑥+5⋅61⋅2𝑥2+5⋅6⋅71⋅2⋅3𝑥3+5⋅6⋅7⋅81⋅2⋅3⋅4𝑥4+⋯ (1−𝑥)−3=1+3𝑥+3⋅41⋅2𝑥2+3⋅4⋅51⋅2⋅3𝑥3+3⋅4⋅5⋅61⋅2⋅3⋅4𝑥4+⋯ be multiplied together, and the coefficient of 𝑥4 in the product be equated to the coefficient of 𝑥4 in the expansion of (1−𝑥)−8, we obtain as the result the sum of the composite series 5⋅6⋅7⋅8×1⋅2+4⋅5⋅6⋅7×2⋅3+3⋅4⋅5⋅6×3⋅4+2⋅3⋅4⋅5×4⋅5+1⋅2⋅3⋅4×5⋅6=4!2⋅11!7!4! 274 Generally, if the given series be 𝑃1𝑄1+𝑃2𝑄2+⋯+𝑃𝑛−1𝑄𝑛−11 where 𝑄𝑟=𝑟(𝑟+1)(𝑟+2)⋯(𝑟+𝑞−1) and 𝑃𝑟=(𝑛−𝑟)(𝑛−𝑟+1)⋯(𝑛−𝑟+𝑝−1) the sum of 𝑛−1 terms will be 𝑝!𝑞!(𝑝+𝑞+1)!(𝑛+𝑝+𝑞−1)!(𝑛−2)!275

Miscellaneous Series

Sum of thepowers of the terms of an Arithmetical Progression 1+2+3+⋯+𝑛=𝑛(𝑛+1)2=𝑆1 1+22+32+⋯+𝑛2=𝑛(𝑛+1)(2𝑛+1)6=𝑆2 1+23+33+⋯+𝑛3=𝑛(𝑛+1)(2𝑛+1)62=𝑆3 1+24+34+⋯+𝑛4=𝑛(𝑛+1)(2𝑛+1)(3𝑛2+3𝑛−1)30=𝑆4 By the method of Indeterminate Coefficients (234). A general formula for the sum of the 𝑟th powers of 1⋅2⋅3⋯𝑛, obtained in the same way is 𝑆𝑟=1𝑟+1𝑛𝑟+1+12𝑛𝑟+𝐴1𝑛𝑟−1+⋯+𝐴𝑟−1𝑛 where 𝐴1, 𝐴2, ⋯ are determined by putting 𝑝=1, 2, 3, ⋯ successively in the equation 12(𝑝+1)! =1(𝑝+2)!+𝐴1𝑟(𝑝)!+𝐴2𝑟(𝑟−1)(𝑝−1)!+⋯+𝐴𝑝𝑟(𝑟−1)⋯(𝑟−𝑝+1)! 276 𝑎𝑚+(a+𝑑)𝑚+(a+2𝑑)𝑚+⋯+(a+𝑛𝑑)𝑚=(𝑛+1)𝑎𝑚+𝑆1𝑚𝑎𝑚−1𝑑+𝑆2𝐶(𝑚,2)𝑎𝑚−2𝑑2+𝑆3𝐶(𝑚,3)𝑎𝑚−3𝑑3+⋯ Proof. By Binomial Theorem and (276).277

Summation of a series partly Arithmetical and partly Geometrical

Example

To find the sum of the series 1+3𝑥+5𝑥2+⋯ to 𝑛 terms. Let 𝑠=1+3𝑥+5𝑥2+7𝑥3++2𝑛−1)𝑥𝑛−1 𝑠𝑥=  𝑥+3𝑥2+5𝑥3++(2𝑛−3)𝑥𝑛−1+(2𝑛−1)𝑥𝑛 ∴ by subtraction, 𝑠(1−𝑥)=1+2𝑥+2𝑥2+2𝑥3+⋯+2𝑥𝑛−1−(2𝑛−1)𝑥𝑛  =1+2𝑥1−𝑥𝑛−11−𝑥−(2𝑛−1)𝑥𝑛 ∴ 𝑠=1−(2𝑛−1)𝑥𝑛1−𝑥+2𝑥(1−𝑥𝑛−1)(1−𝑥)2 278 A general formula for the sum of 𝑛 terms of 𝑎+(a+𝑑)𝑟+(a+2𝑑)𝑟2+(a+3𝑑)𝑟3+⋯ is 𝑆=𝑎−(a+𝑛−1𝑑)𝑟𝑛1−𝑟+𝑑𝑟(1−𝑟𝑛−1)(1+𝑟)2 Obtained as in (278) Rule. Multiply by the ratio and subtract the resulting series. 279 11−𝑥=1+𝑥+𝑥2+𝑥3+⋯+𝑥𝑛−1+𝑥𝑛1−𝑥 280 1(1−𝑥)2=1+2𝑥+3𝑥2+4𝑥3+⋯+𝑛𝑥𝑛−1+(𝑛+1)𝑥𝑛−𝑛𝑥𝑛+1(1−𝑥)2281 (𝑛−1)𝑥+(𝑛−2)𝑥2+(𝑛−3)𝑥3+⋯+2𝑥𝑛−2+𝑥𝑛−1=(𝑛−1)𝑥−𝑛𝑥2+𝑥𝑛+1(1−𝑥)2 By (253) 282 1+𝑛+𝑛(𝑛−1)2!+𝑛(𝑛−1)(𝑛−2)3!+⋯=2𝑛 1−𝑛+𝑛(𝑛−1)2!𝑛(𝑛−1)(𝑛−2)3!+⋯=0 By making 𝑎=𝑏 (125) 283 The series 1-𝑛−32+(𝑛−4)(𝑛−5)3!(𝑛−5)(𝑛−6)(𝑛−7)4!+⋯+(−1)𝑟−1(𝑛−𝑟−1)(𝑛−𝑟−2)⋯(𝑛−2𝑟+1)𝑟! consists of 𝑛2 or 𝑛−12 terms, and the sum is given by 𝑆=3𝑛if 𝑛 be of the form 6𝑚+3, 𝑆=0if 𝑛 be of the form 6𝑚±1, 𝑆=1𝑛if 𝑛 be of the form 6𝑚, 𝑆=2𝑛if 𝑛 be of the form 6𝑚±2, Proof: By (545), putting 𝑝=𝑥+𝑦, 𝑞=𝑥𝑦, and applying (546) 284 The series 𝑛𝑟−𝑛(𝑛−1)𝑟+𝑛(𝑛−1)2!(𝑛−2)𝑟𝑛(𝑛−1)(𝑛−2)3!(𝑛−3)𝑟+⋯ takes the values 0, 𝑛!, 12𝑛(𝑛+1)! according as 𝑟 is <𝑛, =𝑛, or =𝑛+1 Proof: By expanding (𝑒𝑥−1)𝑛, in two ways: first, by the Exponential Theorem and Multinomial; secondly, by the Bin. Th., and each term of the expansion by the Exponential. Equate the coefficients of 𝑥𝑟 in the two results. Other results are obtained by putting 𝑟=𝑛+2, 𝑛+3, ⋯. The series (285), when divided by 𝑟!, is, in fact, equal to the coefficient of 𝑥𝑟 in the expansion of 𝑥+𝑥22!+𝑥33!+⋯𝑛 285 By exactly the same process we may deduce from the function {𝑒𝑥−𝑒−𝑥}𝑛 the result that the series 𝑛𝑟−𝑛(𝑛−2)𝑟+𝑛(𝑛−1)2!(𝑛−4)𝑟−⋯ takes the values 0 or 2𝑛⋅𝑛!, according as 𝑟 is <𝑛 or =𝑛; this series, divided by 𝑟!, being equal to the coefficient of 𝑥𝑟 in the expansion of 2𝑛𝑥+𝑥33!+𝑥55!+⋯𝑛 286

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive

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ID: 210600024 Last Updated: 6/24/2021 Revision: 0 Ref:

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References

  1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science &amp; Engineering
  2. Wheatstone, C., 1854, On the Formation of Powers from Arithmetical Progressions
  3. Stroud, K.A., 2001, Engineering Mathematics
  4. Coolidge, J.L., 1949, The Story of The Binomial Theorem
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