Algebra

Binomial Theorem

(𝑎+𝑏)^𝑛=𝑎^𝑛+𝑛𝑎^𝑛−1𝑏+𝑛(𝑛−1)2!𝑎^𝑛−2𝑏²+𝑛(𝑛−1)(𝑛−2)3!𝑎^𝑛−3𝑏³+⋯

General or (𝑟+1)^th Term

General or (𝑟+1)^th term, 𝑛(𝑛−1)(𝑛−2)⋯(𝑛+𝑟−1)𝑟!𝑎^𝑛−𝑟𝑏^𝑟 or if 𝑛 be a positive integer. 𝑛!(𝑛−1)!𝑟!𝑎^𝑛−𝑟𝑏^𝑟 If 𝑏 be negative, the signs of the even terms will be changed.
If 𝑛 be negative the expansion reduces to (𝑎+𝑏)^−𝑛=𝑎^−𝑛−𝑛𝑎^−𝑛−1𝑏+𝑛(𝑛+1)2!𝑎^−𝑛−2𝑏²−𝑛(𝑛+1)(𝑛+2)3!𝑎^−𝑛−3𝑏³+⋯ General term, (−1)^𝑟𝑛(𝑛−1)(𝑛−2)⋯(𝑛+𝑟−1)𝑟!𝑎^{−𝑛−𝑟}𝑏^𝑟

Euler's Proof

Let the expansionof (1+𝑥)^𝑛 be called 𝑓(𝑛). Then it may be proved by Induction that the equation 𝑓(𝑚)×𝑓(𝑛)=𝑓(𝑚+𝑛) is true when 𝑚 and 𝑛 are integers, and therefore universally true; because the form of an algebraical product is not altered by changing the letters involved into fractional or negative quantities. Hence 𝑓(𝑚+𝑛+𝑝+⋯)=𝑓(𝑚)×𝑓(𝑛)×𝑓(𝑝)×⋯ Put 𝑚=𝑛=𝑝=⋯ to 𝑘 terms, each equal ℎ𝑘, and the theorem is proved for a fractional index.
Again, put −𝑛 for 𝑚, thusm whatever 𝑛 may be, 𝑓(−𝑛)×𝑓(𝑛)=𝑓(0)=1, which proves the theorem for a negative index.
For the greatest term in the expansion of (𝑎+𝑏)^𝑛, take 𝑟 = the integral part of (𝑛+1)𝑏𝑎+𝑏 or (𝑛−1)𝑏𝑎−𝑏, according as 𝑛 is positive or negative.
But if 𝑏 be greater than 𝑎, and 𝑛 negative or fractional, the terms increase without limit.

Examples

Required the 40th term of 1−2𝑥3⁴² Here 𝑟=39; 𝑎=1; 𝑏=−2𝑥3; 𝑛=42. the term will be 42!3!39!−2𝑥3³⁹=−42⋅41⋅401⋅2⋅32𝑥3³⁹

Examples

Required the 31st term of (𝑎−𝑥)⁻⁴ Here, 𝑟=30; 𝑏=−𝑥; 𝑛=−4. the term will be (−1)³⁰4⋅5⋅6⋅⋯⋅30⋅31⋅32⋅331⋅2⋅3⋅⋯⋅30𝑎⁻³⁴(−𝑥)³⁰=31⋅32⋅331⋅2⋅3⋅𝑥³⁰𝑎³⁴

Examples

Required the greatest term in the expansion of 1(1+𝑥)⁶ when 𝑥=1417. And 1(1+𝑥)⁶=(1+𝑥)⁻⁶. Here 𝑛=6, 𝑎=1, 𝑏=𝑥 in the formula (𝑛−1)𝑏𝑎−𝑏=5×14171−1417=2313 therefore 𝑟=23, and the greatest term =(−1)²³5⋅6⋅7⋅⋯⋅271⋅2⋅3⋅⋯⋅231417²³=−24⋅25⋅26⋅271⋅2⋅3⋅41417²³

Examples

Find the first negative term in the expansion of (2𝑎+3𝑏)¹⁷³. Take 𝑟 the first integer which makes 𝑛−𝑟+1 negative; therefore 𝑟>𝑛+1=173+1=623; therefore 𝑟=7. The term will be 173⋅143⋅113⋅83⋅53⋅23(−13)7!⋅(2𝑎)¹³(3𝑏)⁷=−17⋅14⋅11⋅8⋅5⋅2⋅17!⋅𝑏⁷(2𝑎)¹³

Examples

Required the coefficient of 𝑥³⁴ in the expansion of 2+3𝑥2−3𝑥² (2+3𝑥)²(2−3𝑥)²=(2+3𝑥)²(2−3𝑥)⁻²=2+3𝑥2²1−32𝑥⁻² =1+3𝑥+94𝑥²1+23𝑥2+2⋅31⋅23𝑥2²+⋯+333𝑥2³²+343𝑥2³³+353𝑥2³⁴+⋯ the tree terms last written being those which produce 𝑥³⁴ after multiplying (1+3𝑥+94𝑥²); 94𝑥²×333𝑥2³²+3𝑥×343𝑥2³³+1×353𝑥2³⁴ giving for the coefficient of 𝑥³⁴ in the result. 297432³²+10232³³+3532³⁴=30632³² The coefficient of 𝑥^𝑛 will in like manner be 9𝑛32^𝑛−2

Examples

To write the coefficient of 𝑥^3𝑚+1 in the expansion of 𝑥²−1𝑥^22𝑛+1. The general term is (2𝑛+1)!(2𝑛+1−𝑟)!𝑟!𝑥^{2(2𝑛−𝑟+1)}⋅1𝑥^2𝑟=(2𝑛+1)!(2𝑛+1−𝑟)!𝑟!𝑥^{4𝑛−4𝑟+2)} Equate 4𝑛−4𝑟+2 to 3𝑚+1, thus 𝑟=4𝑛−3𝑚+14. Substitute this value of 𝑟 in the general term; the required coefficient becomes (2𝑛+1)![¼(4𝑛+3𝑚+1)]![¼(4𝑛−3𝑚+1)]! The value of 𝑟 shows that there is no term in 𝑥^3𝑚+1 unless 4𝑛−3𝑚+14 is an integer.

Examples

An approximate value of (1+𝑥)^𝑛, when 𝑥 is small, is 1+𝑛𝑥, by neglecting 𝑥² and higher powers of 𝑥.

Examples

An approximation to ³999 by obtaining from the first two or three terms of the expansion of (1000−1)¹²=10−23⋅1000⁻¹³=10−1300=900299300 nearly

Sources and References

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