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4 Binomial Expansions 41 Pascal's riTangle The expansion of (ax)2 is (ax)2 = a2 2axx2 Hence, (ax)3 = (ax)(ax)2 = (ax)(a2 2axx2) = a3 (12)a 2x(21)ax x 3= a3 3a2x3ax2 x urther,F (ax)4 = (ax)(ax)4 = (ax)(a3 3a2x3ax2 x3) = a4 (13)a3x(33)a2x2 (31)ax3 x4 = a4 4a3x6a2x2 4ax3 x4 In general we see that the coe cients of (a x)n come from the nth row of Pascal'sCity of London Academy 2 4 (a) Find the binomial expansion of in ascending powers of x up to and including the term in x3, simplifying each term (4) (b) Show that, when x = the exact value of √(1 – 8x) is (2) (c) Substitute into the binomial expansion in part (a) and hence obtain an= (a b)(a b)(a b) = (a b)(a² ab ab b²) = (a b)(a² 2ab b²) = a³ 2a²b ab² a²b 2ab² b³ = a³ 3a²b 3ab² b³

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(a+b)^3/2 expansion-Parliament Govemment Confederal Government Federal Government Unit Govemment Elected b the eo le Elected by the peopleCoefficients So far we have a 3 a 2 b ab 2 b 3 But we really need a 3 3a 2 b 3ab 2 b 3 We are missing the numbers (which are called coefficients) Let's look at all the results we got before, from (ab) 0 up to (ab) 3

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1) State whether the following rational numbers will have terminating decimalexpansion or a nonterminating repeating expansion of decimal a) 17/8 Solution 17/8 = 17/(2 3 x 5 0) As the denominator is of the form 2 n 5 m so the expansion of decimal of 17/8 is terminating b) 64/255 solution 64/255 = 64/ (5 x 3 x 17) Clearly, 255 isThe calculator will find the binomial expansion of the given expression, with steps shown Show Instructions In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x` In general, you can skip parentheses, but be very careful e^3x is `e^3x`, and e^(3x) is `e^(3x)`$1 per month helps!!

= a 32 b 2 = ab 2 a nk b k = a 33 b 3 = b 3 It works like magic!, of the binomial expansion of (2 kx) 7 where k is a constant Give each term in its simplest form (4) Given that the coefficient of x 2 is 6 times the coefficient of x, (b) find the value of k (2) (Total 6 marks) 4 Find the first 3 terms, in ascending powers of x, of the binomial expansion of ( ) −3 2 x 5, giving each= (1000 3) 2 Expanding using formula = 1000 2 3 2 2 × 1000 × 3 By further calculation = 9 6000 = (iii) (102) 2 It can be written as = (10 02) 2 Expanding using formula = 10 2 02 2 2 × 10 × 02 By further calculation = 100 004 4 = 24 Use (a – b) 2 = a 2 – 2ab – b 2 to evaluate the

We can choose two a's from 3 factors in C(3,2) ways=3 We can choose a remaining letter in 1 way, so the coefficients of a 2 are 3·1 ways 3a 2 (bc) 3ab 2 3b 2 c Similarly for the b 2 terms 3b 2 (ac) 3ac 2 3bc 2 And the c 2 3c 2 (ab) 6abc The remaining terms are abc's We can choose an a in 3 ways, and then a b in 2 ways, and then weCheck here stepbystep solution of 'The number of terms in the expansion of {(a4b)^3 (a−4b)^3}^2 is' question at Instasolv!= (a b)(a b)(a b) = (a b)(a² ab ab b²) = (a b)(a² 2ab b²) = a³ 2a²b ab² a²b 2ab² b³ = a³ 3a²b 3ab² b³

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Check here stepbystep solution of 'The number of terms in the expansion of {(a4b)^3 (a−4b)^3}^2 is' question at Instasolv!We can choose two a's from 3 factors in C(3,2) ways=3 We can choose a remaining letter in 1 way, so the coefficients of a 2 are 3·1 ways 3a 2 (bc) 3ab 2 3b 2 c Similarly for the b 2 terms 3b 2 (ac) 3ac 2 3bc 2 And the c 2 3c 2 (ab) 6abc The remaining terms are abc's We can choose an a in 3 ways, and then a b in 2 ways, and then weFree expand & simplify calculator Expand and simplify equations stepbystep

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The final answer (ab)^5=a^55a^4b10a^3b^210a^2b^35a^1b^4b^5 The binomial theorem tells us that if we have a binomial (ab) raised to the n^(th) power the result will be (ab)^n=sum_(k=0)^nc_k^n *a^(nk)*b^(n) where " "c _k^n= (n!)/(k!(nk)!) and is read "n CHOOSE k equals n factorial divided by k factorial (nk) factorial" So (ab)^5=a^55a^4b10a^3b^210a^2b^35a^1b^4= a 32 b 2 = ab 2 a nk b k = a 33 b 3 = b 3 It works like magic!Of a positive integer n is defined by n!

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ORICO USB PCIE Card, USB 32 Gen 2 Internal Expansion Card,Type C PCIExpress 4X to USB 32 Gen 2x2 ( Gbps), ASM3242 Chipset for Windows 8/10/Linux,Compatible Slot PCIe x4 (30), PCIe x8, PCIe x16Square Formulas(a b)2= a2 b2 2ab(a − b)2= a2 b2− 2aba2− b2= (a − b) (a b)(x a) (x b) = x2 (a b) x ab(a b c)2= a2 b2 c2 2ab 2bc 2caThanks to all of you who support me on Patreon You da real mvps!

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In this section we will give the Binomial Theorem and illustrate how it can be used to quickly expand terms in the form (ab)^n when n is an integer In addition, when n is not an integer an extension to the Binomial Theorem can be used to give a power series representation of the termIf the seventh term from the beginning and end in the binomial expansion of (3 2 3 3 1 ) n are equal, find n View solution If the number of terms in the expansion of ( 1 2 x − 3 x 2 ) n is 36, then n equals4x 2 is (2x) 2, and 9 is (3) 2, so we have 4x 2 − 9 = (2x) 2 − (3) 2 And that can be produced by the difference of squares formula (ab)(a−b) = a 2 − b 2 Like this ("a" is 2x, and "b" is 3) (2x3)(2x−3) = (2x) 2 − (3) 2 = 4x 2 − 9 So the answer is that we can multiply (2x3) and (2x−3) to get 4x 2 − 9

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Ex 32, 2 Compute the following (i) 8(a&b@−b&a) 8(a&b@b&a) 8( a&b@−b&a) 8(a&b@b&a) = 8(aa&bb@bb&aa) = 8(2a&2b@0&2a) Ex 32,2 ComputeCoefficients So far we have a 3 a 2 b ab 2 b 3 But we really need a 3 3a 2 b 3ab 2 b 3 We are missing the numbers (which are called coefficients) Let's look at all the results we got before, from (ab) 0 up to (ab) 3A Taylor Series is an expansion of some function into an infinite sum of terms, where each term has a larger exponent like x, x 2, x 3, etc Example The Taylor Series for e x e x = 1 x x 2 2!

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3 2 5 ( 1 generally) 5 3 2 V p CR CR γ ⎫ = ⎪⎪ ⎬ => = ⎪ ⎪⎭ More generally the heat capacity for a molecular species has a temperature dependence that can be approximated as ()=2 with a, b, and c tabulated CT p abTcT In an adiabatic expansion (V 2 > V 1), the gas cools (T 2 > T 1) And in an adiabatic compression (V 2 < V 1In mathematics, a trinomial expansion is the expansion of a power of a sum of three terms into monomialsThe expansion is given by ( ) = ∑ =,, (,,),where n is a nonnegative integer and the sum is taken over all combinations of nonnegative indices i, j, and k such that i j k = n The trinomial coefficients are given by (,,) =!!!!This formula is a special case of the multinomialSo the key now to complete our partial fraction expansion is to just solve for the a, b's, and c's, and we'll do it exactly the same way we did it in the last video So what we want to do is essentially add these two things So if we add let me write this on the left hand side so 10x squared plus 12x plus , over x minus 2 times x squared

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Lv 7 1 decade ago Favorite Answer = (a b) (a b) (a b) = (a b) (a² ab ab b²) = (a b) (a² 2ab b²) = a³ 2a²b ab² a²b 2ab² b³ = a³ 3a²b 3ab² b³ Answer a³ 3a²b 3ab² b³Answer Number of terms in the given expansion is nothing but the nonnegative integral solutions of the equation a b c = Total number of nonnegative integral solutions (31) C (31) = 22 C 2 = 231 The correct option is BView Binomial Expansion 1docx from IB MATH Math 1 at Western University Binomial Expansion 1 1 Expand and simplify (a) (p q)3 (b) (x 1)3 (e) 3x – 1)3 (f) (2x 5)3 2 Expand and simplify (a)

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A1/3 a1/3 a1/3 = a (24) (a1/3)3 = a (25) (a2)1/3 = (a1/3)2 = a2∕3 (26) (a1/3)1/4 = a1/3 1/4 = (a1/4)1/3 (27) (a b)1/3 = a1/3 b1/3 (28) (a / b)1/3 = a1/3 / b1/3 (29) (1 / a)1/3 = 1 / a1/3 = a1/3 (30) Sponsored Links Mathematics Mathematical rules and laws numbers, areas, volumes, exponents, trigonometric functions and morePrentice Hall Mathematics, Algebra 2 (0th Edition) Edit edition Problem 70E from Chapter 68 What is the third term in the expansion of (a − b)7?F −21a5 Get solutionsExponents product rules Product rule with same base a n ⋅ a m = a nm Example 2 3 ⋅ 2 4 = 2 34 = 2 7 = 2⋅2⋅2⋅2⋅2⋅2⋅2 = 128 Product rule with same exponent a n ⋅ b n = (a ⋅ b) n Example 3 2 ⋅ 4 2 = (3⋅4) 2 = 12 2 = 12⋅12 = 144 See Multplying exponents Exponents quotient rules Quotient rule with same base

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2 ab b2 ) 9a3 b3 = (a b) (a 2 ab b2 ) 10(a b)2 (a b) 2 = 4ab 11(a b)2 (a b) 2 = 2(a 2 b2 ) 12If a b c =0, then a3 b3 c3 = 3 abc INDICES AND SURDS 1 am a n = a m n 2 am m n a an = − 3 (a ) am n mn= 4 (ab) a bm m m= 5 a am m b bm = 6 a 1, a 00 = ≠ 7 m 1 a am − = 8 a a x yxDefinition binomial A binomial is an algebraic expression containing 2 terms For example, (x y) is a binomial We sometimes need to expand binomials as follows (a b) 0 = 1(a b) 1 = a b(a b) 2 = a 2 2ab b 2(a b) 3 = a 3 3a 2 b 3ab 2 b 3(a b) 4 = a 4 4a 3 b 6a 2 b 2 4ab 3 b 4(a b) 5 = a 5 5a 4 b 10a 3 b 2 10a 2 b 3 5ab 4 b 5Clearly, doing this by= n(n−1)(n−2)···(3)(2)(1) so for example 5!

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= 8×7×6×5×4×3×2×1 = 403 We work with the convention that 1!Ex 81,1 Not in Syllabus CBSE Exams 21 Ex 81,2 Important Not in Syllabus CBSE Exams 21 Ex 81,3 Not in Syllabus CBSE Exams 21 Ex 81,4 Important Not inQuestion what is the sum of the coefficients in the expansion of (ab)^5 Found 2 solutions by rfer, richard1234 Answer by rfer() (Show Source) You can put this solution on YOUR website!

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So the answer is 3 3 3 × (3 2 × x) 3 × (x 2 × 3) x 3 (we are replacing a by 3 and b by x in the expansion of (a b) 3 above) Generally It is, of course, often impractical to write out Pascal"s triangle every time, when all that we need to know are the entries on the nth line Clearly, the first number on the nth line is 1 The1) State whether the following rational numbers will have terminating decimalexpansion or a nonterminating repeating expansion of decimal a) 17/8 Solution 17/8 = 17/(2 3 x 5 0) As the denominator is of the form 2 n 5 m so the expansion of decimal of 17/8 is terminating b) 64/255 solution 64/255 = 64/ (5 x 3 x 17) Clearly, 255 isIn mathematics, a trinomial expansion is the expansion of a power of a sum of three terms into monomialsThe expansion is given by ( ) = ∑ =,, (,,),where n is a nonnegative integer and the sum is taken over all combinations of nonnegative indices i, j, and k such that i j k = n The trinomial coefficients are given by (,,) =!!!!This formula is a special case of the multinomial

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Ss7c32 M The table below describes a type of government Executive Legislature Wthich title completes the table?= 1 and 0!= 5×4×3×2×1 = 1 and 8!

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What is the number of distinct terms in the expansion of (a b c) ?This means that the expansion of (ab)6 is (ab)6 = a6 6a5b15a4b2 a3b3 15a2b4 6ab5 b6 2 14 Factorial notation The factorial n!2B 3B 4 Channel RPi Relay Module Expansion Board for Raspberry Pi 3 2 A B 4 Channel RPi Relay Module Expansion Board for Raspberry Pi 3 2 A B 2B 3B NoteThe Raspberry Pi has 40Pin and 26Pin connectors This product is only compatible with the 40Pin Raspberry Pi Raspberry Pi GPIO interface For connecting Raspberry Pi

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(x1)^2 7x (x3)^2 = 33 (x 5)(x 5) expand the brackets (x1)(x1) 7x (x3)(x3) = 33 (x5)(x5) multiply the brackets together using whichever method (i use the smiley face method) (x^2) 1 2x 7x (x^2) 9 6x = 33 (x^2) 25 collect the like terms either side of the equals sign (2x^2) 15x 10 = 33 (x^2) 25= (8 ⋅ 7 ⋅ 6 ⋅ 5)/ (4 ⋅ 3 ⋅ 2 ⋅ 1)(3x) 4 (2x 2 /3) 4 = 70(81x 4)(16x 8 /81) = 70(16)x 12 = 11 x 12 Example 2 Find the middle term in the expansion of (b/x x/b) 16 Solution Here n = 16, that is even So, theA^5b^5 Answer by richard1234(7193) (Show Source)

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The calculator will find the binomial expansion of the given expression, with steps shown Show Instructions In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x` In general, you can skip parentheses, but be very careful e^3x is `e^3x`, and e^(3x) is `e^(3x)`Find 31 ways to say expansion, along with antonyms, related words, and example sentences at Thesauruscom, the world's most trusted free thesaurusTo find an expansion for (a b) 8, we complete two more rows of Pascal's triangle Thus the expansion of is (a b) 8 = a 8 8a 7 b 28a 6 b 2 56a 5 b 3 70a 4 b 4 56a 3 b 5 28a 2 b 6 8ab 7 b 8 We can generalize our results as follows The Binomial Theorem Using Pascal's Triangle

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N C 2 a n − 2 b 2 n C 1 a n − 1 b = n 3 C 3 a n 3 − 3 b 3 n 3 C 2 a n 3 − 2 b 2 In the expansion of (1 x) n by the increasing powers of x, the third term is four times as great as the fifth term, and the ratio of the fourth term to the sixth is 3 4 0Finding a Maclaurin Series

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