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Mathematics Dictionary
Dr. K. G. Shih
Binomial Theorem and Pascal Triangle
Questions

  • Q01 | - Defintion of Binomial Theorem
  • Q02 | - Coefficients of Binomial expansaion
  • Q03 | - Pascal triangle and coefficients of Binomial expansaion
  • Q04 | - Sequence along r = 2 column : 1, 3, 06, 10, 15, ......
  • Q05 | - Sequence along r = 3 column : 1, 4, 10, 20, 35, ......
  • Q06 | - Sequence along r = 4 column : 1, 5, 15, 35, 70, .....
  • Q07 | - Sequence along r = 5 column : 1, 6, 21, 56, 126 ....
  • Q08 | - Find Sum[n^4] usimg Sum[C(n+3,4)] = C(n+4,5)
  • Q09 | - Fibonacci's sequence in Pascal triangle
  • Q10 | - Formula
  • Q11 | - Quiz for binomial theorem
  • Q12 | - Answers for Quiz for binomial theorem
  • Q13 | - Use Pascal triangle find coefficients of (x+y)^7
  • Q14 | - Expand (x+1)^n
  • Q15 | - Find coefficient (x^3)*(y^5) in expansion of (x+y)^n
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, 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) + ...
  • 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

Q3. Coefficients of binomial expansion and Pascal triangle
A3. 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
  • 1, 5, 10, 10, 5, 1 are coefficients of expansion of (x+y)^5
[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 power n = 3 + 5 = 8
  • For y^5 we know r = 5
  • At row n = 8 and r = 5 we have C(n,r) = C(8,5) = 70
Go to Begin

Q4. What is r2 series ?
A4. Answer
 [Example] Prove that Sum[n*(n+1)/2] = n*(n+1)*(n+2)/3!
  • Sum[n*(n+1)/2]
  • = Sum[n^2/2]+Sum[n/2]
  • = (Sum(n^2) + Sum[n])/2
  • = (n*(n+1)*(2*n+1)/6 + n*(n+1)/2)/2
  • = n*(n+1)*(2*n+1)/12 + n*(n+1)/4
  • = n*(n+1)*(2*n+1)/3+1)/4
  • = n*(n+1)*(2*n+4)/12
  • = n*(n+1)*(n+2)/3! where 3!=1*2*3=6
  • = C(n+2,3) the coefficients of binomial expansion
  • Hence Sum[C(n+1,2)] = C(n+2,3)
Go to Begin

Q5. What is r3 series ?
A5. Answer
  • Squence along r3 column : 1, 4, 10, 20, 35, 56, ....
  • T(n) = nth term = n*(n+1)*(n+2)/3!
  • S(n) = Sum[n*(n+1)*(n+2)/3!] = n*(n+1)*(n+2)*(n+3)/4!
  • Hence Sum[C(n+2, 3)] = C(n+3, 4)
    • 1st difference is r2 series
    • 2nd difference is r1 series
    • 3nd difference is common difference d=1
[Example] Prove that Sum[n*(n+1)*(n+2)/3!] = n*(n+1)*(n+2)*(n+3)/4!
  • Sum[n*(n+1)*(n+2)/6]
  • = Sum[n^3/6]+Sum[3*n^2/6]+Sum[2*n/6]
  • = (n*(n+1)/2)/6 + n*(n+1)*(2*n+1)/12 + n*(n+1)/12
  • = n*(n+1)*[(n*(n+1)/24 + (2*n+1)/12 +1/12]
  • = n*(n+1)*(n^2+5*n+6)/24
  • = n*(n+1)*(n+2)*(n+3)/4! where 4!=1*2*3*4
  • Hence Sum[C(n+2,3)] = C(n+3,4)
Go to Begin

Q6. What is the sequence along r4 column ?
A6. Answer
  • Sequence along r4 column : 1, 5, 15, 35, 70, ....
  • T(n) = nth term = n*(n+1)*(n+2)*(n+3)/4!
  • S(n) = Sum[T(n)] = n*(n+1)*(n+2)*(n+3)*(n+4)/5!
  • Hence Sum[C(n+3, 4)] = C(n+4, 5)
  • Difference
    • 1st difference is r3 series
    • 2nd difference is r2 series
    • 3rd difference is r1 series
    • 4th difference is common difference d=1
Go to Begin

Q7. What is the sequence along r5 column ?
A7. Answer
[Definition]
  • Sequence along r5 column
  • T(n) = nth term = n*(n+1)*(n+2)*(n+3)(n+4)/5!
  • S(n) = Sum[T(n)] = n*(n+1)*(n+2)*(n+3)*(n+4)/5!
  • Hence Sum[C(n+4, 5)] = C(n+5, 6)
  • Difference
    • 1st difference is r4 series
    • 2nd difference is r3 series
    • 3rd difference is r2 series
    • 4th difference is r1 series
    • 5th difference is common difference d=1
[Coclusion]
  • Su,[C(n+0,1)] = C(n+1,2)
  • 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)
Go to Begin

Q8. Find Sum[n^4] by using Sum[C(n+3,4)] = C(n+4,5)
A8. Answer
  • Sum[C(n+3),4] = Sum[n*(n+1)*(n+2)*(n+3)/4!] = C(n+4,5)
  • (Sum[n^4] + 6*Sum[n^3] + 11*Sum[n^2] + 6*Sum[n])/24 = C(n+4,5)
  • We know that
    • Sum[1] = n.
    • Sum[n] = n*(n+1)/2.
    • Sum[n^2] = n*(n+1)*(2*n+1)/6.
    • Sum[n^3] = (n*(n+1)/2)^2
  • The answer is Sum[n^4] = (6*n^5 + 5*n^4 + 10*n^3 - n)/30
Go to Begin

Q9. Fibonacci's sequence in Pascal triangle
A5. Answer in Fibonacc's pattern
  • Fibonacci's Sequence : 1, 1, 2, 3, 5, 8, 13, ...
  • Recursion formula :
    • T(0) = 0.
    • T(1) = 1.
    • T(n+2) = T(n+1) + T(n)
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. Quiz for binomial theorem

  • 1. Expand (x + y)^5.
  • 2. Find cosntant term in expansion of (x + 1/x)^10.
  • 3. Find coefficients of (x^5)*(y^5) in expansion of (x + y)^n.
  • 4. Is there a costant term in (x^2 + 1/x)^20 ?
  • 5. How many trailor zeros in 25! ?
  • 6. Find Sqr(2) by using the expansion of Sqr(1 + x) to 3 decimal place.
  • 7. Using Pascal triangle, write down the coefficients of (x+y)^7.
  • 8. Find the coefficients of (x)*(y^99) in expansion of (x+y)^n.
  • 9. Express 1/(1+x) as seires.
  • 10 Find C(100,98).
Go to Begin

Q12. Quiz for binomial theorem

  • 1. Expand (x + y)^5.
    • Using Pascal triangle, we have
    • (x+y)^5 =x^5+ 5*(x^4)*y+ 10*(x^3)*(x^2)+ 10*(x^2)*(y^3)+ 5*x*(y^4)+ y^5
    • Pascal triangle is faster than using binomial theorem.
  • 2. Find cosntant term in expansion of (x + 1/x)^10.
    • Let the term be C(n,r)*(x^(n-r))*((1/x)^r).
    • Where n = 10, we have to find r.
    • x power + y power = (n - r) + (-r) = 0 if it is constant.
    • Hence r = 5.
    • Coefficient = C(10,5) = (10*9*8*7*6)/(5*4*3*2*1) = 252.
  • 3. Find coefficients of (x^5)*(y^5) in expansion of (x + y)^n.
    • x power + y power = 5 + 5 = 10
    • Coefficient = C(10,5) = (10*9*8*7*6)/(5*4*3*2*1) = 252.
  • 4. Is there a costant term in (x^2 + 1/x)^20 ?
    • C(n,r)*((x^2)^(n-r))*((1/x)^r) = C(n,r)*(x^(2*n-2*r))*(x^(-r))
    • Since the term is constant, the power of x is 2*n - 3*r = 0
    • Now n = 20 and r = 40/3. Hence there is no constant term.
  • 5. How many trailor zeros in 25! ?
    • Trailor zeros = 25/5 + 25/25 + 25/125 + ....
    • Since each tem is integer.
    • Hence trialor zeros = 5 + 1 + 0 = 6.
  • 6. Find Sqr(2) by using the expansion of Sqr(1 + x) to 1 decimal place.
    • Sqr(1+x) = (1+x)^(1/2)
    • Sqr(1+x) = 1+ (1/2)*x+ (1/2)(-1/2)*x^2/2!+ (1/2)*(-1/2)*(-3/2)*x^3/3! ...
    • Sqr(1+x) = 1 + x/2 - x^2/((2^2)*(2!)) + 3*x^3/((2^3)*(3!))
    • - (1*3*5)*x^4/((2^4)*(4!)) + (1*3*5*7)*x^5/((2^5)*(5!)) + ...
    • Sqr(1+1) = 1 + 1/2 - 1/8 + 1/16 - 15*1/(16*24) + ....
    • Sqr(2) = 1 + 0.5 - 0.125 + 0.0625 - 0.0390625 + ...
    • Sqr(2) = 1.3984375 + 245/3840 - ...... (converge very slow).
  • 7. Using Pascal triangle, write down the coefficients of (x+y)^7.
    • 1, 7, 21, 35, 35, 21, 7, 1.
  • 8. Find the coefficients of (x)*(y^99) in expansion of (x+y)^n.
  • 9. Express 1/(1+x) as seires.
    • 1/(1+x) = (1+x)^(-1)
    • (1+x)^(-1) =1 + (-1)*x+ (-1)*(-1-1)*x^2/2!+ (-1)*(-2)*(-3)*x^3/3! + ...
    • (1+x)^(-1) =1 - x + x^2 - x^3 + .....
  • 10 Find C(100,98).
    • C(100,98) = C(100,2) = 100*(100-1)/2! = 4950.
Go to Begin

Q13. Use Pascal triangle write down coefficient of expansion terms of (x+y)^7
A13. 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.
  • Hence the answer is
    • 1 ........................... for (x^7)*(y^0)
    • 1 + 6 = 7 ................... fof (x^6)*(y^1)
    • 6 + 15 = 21 ................. for (x^5)*(y^2)
    • 15 + 20 = 35 ................ for (x^4)*(y^3)
    • 20 + 15 = 35 ................ for (x^3)*(y^4)
    • 15 + 06 = 21 ................ for (x^2)*(y^5)
    • 06 + 01 = 07 ................ for (x^1)*(y^6)
    • 1 ........................... for (x^0)*(y^7)
Example : 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.
Go to Begin

Q14. Expansion of (x+1)^n
A14. 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.
  • (x+1)^5 = x^5 + 5*x^4 + 10*x^3 + 10*x^2 + 5*x + 1
Notes
  • x^0 = 1.
  • x^1 = x.
Go to Begin

Q15. Find coefficient 0f (x^3)*(y^5) in expansion (x+y)^n
A15. 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.
  • Or use formula : C(8,5) = C(8,3) = (8*7*6)/(3*2*1) = 56.
Go to Begin
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