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Mathematics Dictionary
Dr. K. G. Shih
Binomial Theorem and Calculus
Questions

  • Q01 | - What is Binomial Theorem ?
  • Q02 | - Coefficients of Binomial expansaion
  • Q03 | - Express 1/(1+x^2) into series and find series of arctan(x)
  • Q04 | - How to get the binomial series ?
  • Q05 | - Express 1/Asq(1-x^2) in series and Find seires of arcsin(x)
  • Q06 | - Expand x/(1-x)^2 in series and find Sum[n/(2^n)]
  • Q07 | - Express Sqr(1+x) in series
  • Q08 | - Express 1/(1+x) in series
  • Q09 | -
  • Q10 | - Formula
  • Q11 | - Pascal triangle and binomial theorem
Answers


Q1. Deifintion of Binomial Theorem
A1. Answer
  • (x + y)^n = C(n,0)*x^n + C(n,1)*(x^(n-1))*(y)+ C(n,2)*(x^(n-2))*(y^2) + ....
    • C(n,r) = n*(n-1)*(n-2)*...*(n-r+1)/r!.
    • C(n,0) = C(n,n) = 1.
    • C(n,1) = C(n,n-1) = n
  • It can be expressed as (x+y)^n = Sum[C(n,r)*(x^(n-r))*(y^r)]
  • It has (n + 1) terms.
  • Sum of the coeefficients = C(n,0) + C(n,1) + .... = 2^n.
    • Let x = y = 1 and we have.
    • C(n,0) + C(n,1) + .... = 2^n.
  • Sum of the coeefficients of odd terms = sum of coefficients of even terms.
    • Let x = 1 and y = -1, we have
    • C(n,0) + C(n,2) + ... = C(n,1) + C(n,3) + ...
  • Prove that C(n,r) = C(n,n-r)
    • C(n,r) = n*(n-1)*...*(n-r+1)/r!
    • = n*(n-1)*(n-2)*...*(n-r+1)*(n-r)!/((r!)*(n-r)!)
    • = n!/((r!)*(n-1)!)
    • C(n,n-r) = n*(n-1)*(n-2)*...*(n-(n-r)+1)/(n-1)!
    • = n*(n-1)*(n-2)*....*(r+1)/(n-1)!
    • = n*(n-1)*(n-2)*....*(r+1)*r!/((n-1)!)*r!)
    • = n!/((r!)*(n-1)!)
    • Hence C(n,r) = C(n,n-r).
Go to Begin

Q2. Coefficients of Binomial expansion
A2. Answer
  • C(n,r) is the coefficient (x^(n-r))*(y^r) in (x+y)^n
  • x power + y power = n.
  • Find coefficient of (x^5)*(x*4) in expansion of (x+y)^n.
    • n = 5 + 4 and r = 4.
    • Coeficient of (x^5)*(y^4) = C(9,4)
    • = 9*8*7*(9-4+1)/4!
    • = (9*8*7*6)/(4*3*2*1)
    • = 126.
Go to Begin

Q03. Application : Express 1/(1+x^2) into series and find series of arctan(x)

Use binomial rheorem
  • n = -1
  • (1 + x^2)^(-1) = 1 + (-1)*(x^2) + (-1)*(-2)*((x^2)^2)/2! + ....
  • (1 + x^2)^(-1) = 1 - x^2 + x^4 - x^6 + x^8 - ....
Example : Find series of arctan(x)
  • Arctan(x) = ∫dx/(1+x^2)
  • Arctan(x) = ∫(1 - x^2 + x^4 - x^5 + .....)*dx
  • Arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + .....
Go to Begin

Q04. How to get the binomial series

Use Newton's accomplishments
  • F (x) = (1 + x)^n .................................. F (0) = 1
  • F'(x) = n*(1 + X)^(n-1) ............................ F'(0) = n
  • F"(x) = n*(n-1)*(1 - x)^(n-2) ...................... F"(0) = n*(n-1)
  • Similarly, find 3rd derivative, 4th derivative, ...
  • Hence (1+x)^n = F(0)/0! + F'(0)*x/1! + F"(0)*x^2/2! + ....
  • Hence (1+x)^n = 1 + n*x + n*(n-1)*x^2/2! + .....
Go to Begin

Q05. Express 1/Asq(1-x^2) in series and Find seires of arcsin(x)

  • 1/Asq(1 - x^2) = (1 - x^2)^(-1/2).
  • = 1+ (-1/2)*(-x^2)+ (-1/2)*(-1/2-1)*(-x^2)^2+ (-1/2)*(-1/2-1)*(-1/2-2)*(-x^2)^3
  • = 1 + x^2 + (1*3)*x^4/(2^2) + (1*3*5)*x^6/(2^3) + .....
  • Find series of arcsin(x)
    • Since arcsin(x) = ∫dx/Abs(1-x^2)
    • arcsin(x) = x + x^3/3 + (1*3)*x^5/(4*2^2) + (1*3*5)*x^7/(6*2^3) + ....
    • arcsin(x) = x + x^3/3 + 3*x^5/16 + 5*x^7/24 + ......
Go to Begin

Q06. Expand x/(1-x)^2 in series and find Sum[n/(2^n)]

  • x/(1-x)^2 = x*(1-x)^(-2).
  • = x*(1 + (-2)*(-x) + (-2)*(-3)*(-x)^2/2! + (-2)*(-3)*(-4)*(-x)^3/3! +....
  • = x*(1 + 2*x + 3*x^2 + 4*x^3 + ....)
  • = x + 2*x^2 + 3*x^3 + 4*x^4 + .....
  • Hence x/(1-x)^2 = x + 2*x^2 + 3*x^3 + 4*x^4 + .....
  • Find Sum[n/(2^n)]
    • Let x = 1/2,
    • Then left hand side = x/(1-x)^2 = (1/2)/(1-1/2)^2 = 2.
    • Then right hand side = 1/2 + 2/2^2 + 3/2^3 + 4/2^4 + ... = Sum[n/(2^n)]
    • Hence Sum[n/(2^n)] = 2.
Go to Begin

Q07 Express Sqr(1+x) in series
  • Sqr(1+x) = (1+x)^(1/2).
  • (1+x)^(1/2) = 1+(1/2)*x+(1/2)*(-1/2)*x^2/2!+(1/2)*(-1/2)*(-3/2)*x^3/3!+.....
  • (1+x)^(1/2) = 1 + x/2 - x^2/((2!)*(2^2)) + (1*3)*x^3/((3!)*(2^3)) + .....
  • (n+1)th term = ((-1)^n)*(1*3*5*....*(2*n-3))/((n!)*(2^n))
Go to Begin

Q08 Express 1/(1+x) in series
  • 1/(1+x) = (1+x)^(-1).
  • (1+x)^(-1) = 1 + (-1)*x +(-1)*(-2)*x^2/2! + (-1)*(-2)*(-3)*x^3/3! +.....
  • (1+x)^(1/2) = 1 - x + x^2 - x^3 + .....
Go to Begin

Q09 Go to Begin

Q10. Formula

[Sum in C(n,r) Form]
  • Sum[C(n+1,2)] = C(n+2,3)
  • Sum[C(n+2,3)] = C(n+3,4)
  • Sum[C(n+3,4)] = C(n+4,5)
  • Sum[C(n+4,5)] = C(n+5,6)
[Sum in factor Form]
  • Sum[1] = n
  • Sum[n] = n*(n+1)/2!
  • Sum[n*(n+1)/2!] = n*(n+1)*(n+2)/3!
  • Sum[n*(n+1)*(n+2)/3!] = n*(n+1)*(n+2)*(n+3)/4!
  • Sum[n*(n+1)*(n+2)*(n+3)/4!] = n*(n+1)*(n+2)*(n+3)*(n+4)/5!
[Other series] Go to Begin

Q11. Pascal triangle and binomial theorem
A11. Answer
  • 01 .......................... Coeff of expansion terms of (x+y)^0.
  • 01 01 ....................... Coeff of expansion terms of (x+y)^1.
  • 01 02 01 .................... Coeff of expansion terms of (x+y)^2.
  • 01 03 03 01 ................. Coeff of expansion terms of (x+y)^3.
  • 01 04 06 04 01 .............. Coeff of expansion terms of (x+y)^4.
  • 01 05 10 10 05 01 ........... Coeff of expansion terms of (x+y)^5.
  • 01 06 15 20 15 06 01 ........ Coeff of expansion terms of (x+y)^6.
How to use Pascal triangle to expand (x+y)^4 ?
  • 1st term is 1*(x^4)*(y^0) and x power + y power = 4 + 0 = 4.
  • 2nd term is 4*(x^3)*(y^1) and x power + y power = 3 + 1 = 4.
  • 3rd term is 6*(x^2)*(y^2) and x power + y power = 2 + 2 = 4.
  • 4th term is 4*(x^1)*(y^3) and x power + y power = 1 + 3 = 4.
  • 5th term is 1*(x^0)*(y^4) and x power + y power = 0 + 4 = 4.
Properties of (x+y)^4
  • It expansion has 4 + 1 = 5 terms.
  • x power + y power = 4.
Reference
Go to Begin

Q16. Coefficients of Binomial expansion
A16. Expansions of (x+1)^n
  • (x+1)^1 = x^1 + 1*x^0
  • (x+1)^2 = x^2 + 2*x^1 + 1.
  • (x+1)^3 = x^3 + 3*x^2 + 3*x + 1.
  • (x+1)^4 = x^4 + 4*x^3 + 6*x^2 + 4*x + 1.
Notes
  • x^0 = 1.
  • x^1 = x.
Go to Begin

Q17. Coefficients of binomial expansion and Pascal triangle
A17. Answer
[Defintion] Pascal triangle : It is coeff of binomial expansion
  • 1, 2, 1 are ceeficients of expansion of (x+y)^2
  • 1, 3, 3, 1 are ceeficients of expansion of (x+y)^3
  • 1, 4, 6, 4, 1 are ceeficients of expansion of (x+y)^4
[Pscal triangle]
    Column ...... r0 r1 r2 r3 r4 r5 r6 r7 r8
    ---------------------------------
    n=0 ......... 01
    n=1 ......... 01 01
    n=2 ......... 01 02 01
    n=3 ......... 01 03 03 01
    n=4 ......... 01 04 06 04 01
    n=5 ......... 01 05 10 10 05 01
    n=6 ......... 01 06 15 20 15 06 01
    n=7 ......... 01 07 21 35 35 21 07 01
    n=8 ......... 01 08 28 56 70 56 28 08 01
[Example] Find coefficient of (x^3)*(y^5) in expansion of (x+y)^n
  • The x power and y power = n = 3 + 5 = 8.
  • For y^5 we know it is in (5+1)th term (r =5).
  • At row n=8 and r=5 we have C(n,r) = C(8,5) = 56.
  • Hence the coefficient of (x^3)*(y^5) is 56.
Go to Begin
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