Mathematics Dictionary

Mathematics Dictionary
Dr. K. G. Shih

Binomial Theorem and Pascal Triangle

Questions

AL 07a 01 | - C(n,r) + C(n,r+1) = C(n+1,r+1)
AL 07a 02 | -
AL 07a 03 | -
AL 07a 04 | -
AL 07a 05 | -
AL 07a 06 | -
AL 07a 07 | -
AL 07a 08 | -
AL 07a 09 | -
AL 07a 10 | -
AL 07a 11 | -
AL 07a 12 | -
AL 07a 13 | - Use Pascal triangle write down coefficient of expansion terms of (x+y)^7
AL 07a 14 | -
AL 07a 15 | -
AL 07a 16 | -
AL 07a 17 | - Quiz for binomial theorem
AL 07a 18 | - Answers for Quiz for binomial theorem

Answers

AL 07 01. C(n,r) + C(n,r+1) = C(n+1,r+1)

Proof for n = 1,2,3

The proof is given in AL 07 17

Proof for n = n+1

Step 1

(a+x)^(n+1) = (a+x)*((a+x)^n)
= (a+x)*Sum[C(n,r)*(a^(n-r))*(x^r))]
= Sum1[C(n,r)*(a^(n-r+1))*(x^r))] + Sum2[C(n,r)*(a^(n-r))*(x^(r+1))]
Where summation is from r = 0 to r = n

Step 2 : Change (a^(n-r))*(x^(r+1)) to (a^(n-r+1))*(x^r))

The 2nd summamtion in step 1 can be
Sum2[C(n,r-1)*(a^(n-r+1))*(x^r)] for r = 1 to r = n+1
(a+x)^(n+1) = Sum1[C(n,r)*(a^(n-r+1))*(x^r))] + Sum2[C(n,r-1)*(a^(n-r+1))*(x^r)]
Sum1 from r = 0 to r = n and Sum2 from r = 1 to r = n+1

Step 3 : Change Sum1 : r=0 to r = n as r = 1 to r = n

Sum1[C(n,r)*(a^(n-r+1))*(x^r))]
= C(n,0)*(a^(n+1))*(x^0) + Sum1[C(n,r)*(a^(n-r+1))*(x^r))]

Step 4 : Change Sum2 : r=1 to r = n+1 as r = 1 to r = n

Sum2[C(n,r-1)*(a^(n-r+1))*(x^r))]
= Sum2[C(n,r-1)*(a^(n-r+1))*(x^r))] + C(n,n)*(a^0)*(x^(n+1))

Step 4 : Put Sum1 and Sum2 together
- Step 5
  - (a+x)^(n+1) = C(n+1,0)*(a^(n+1))*(x^0) + Sum1[C(n,r)*(a^(n-r+1))*(x^r))]
  - + Sum2[C(n,r-1)*(a^(n-r+1))*(x^r))] + C(n+1,n+1)*(a^0)*(x^(n+1))
  - Sum1 and Sum2 both from r = 1 to n
  - (1+x)^(n+1) = Sum[C(n+1,r)*(a^(n-r+1))*(x^(r))] for limit r=0 to r=n+1
  - Hence C(n+1,r) = C(n,r-1) + C(n,r)
  Go to Begin
  AL 07 02. Coefficients of Binomial expansion
  
  Expand (x+y)^2
  
  (x+y)^2 = x^2 + 2*x*y + y^2
  Coefficients : 1, 2, 1
  Expand (x+y)^3
  
  (x+y)^3 = x^3 + 3*(x^2)*y + 3*x*y^2 + y^3
  Coefficients : 1, 3, 3, 1
  Expand (x+y)^4
  
  (x+y)^4 = x^4 + 4*(x^3)*y + 6*(x^2)*(y^2) + 4*x*y^3 + y^4
  Coefficients : 1, 4, 6, 4, 1
  Example : Find coefficients in (x + y)^4 from coefficients in expansion of (x+y)^3
  
  Coefficients in (x+y)^3 : C(3,0)=1, C(3,1)=3, C(3,2)=3, C(3,3)=1
  Coefficients in (x+y)^4 : 1, 4, 6, 4, 1
  Coefficients of 1st term in (x+Y)^4 = C(4,0) = C(3,0) = ................ 1
  Coefficients of 2nd term in (x+Y)^4 = C(4,1) = C(3,0) + C(3,1) = 1 + 3 = 4
  Coefficients of 3rd term in (x+Y)^4 = C(4,2) = C(3,1) + C(3,2) = 3 + 3 = 6
  Coefficients of 4th term in (x+Y)^4 = C(4,3) = C(3,2) + C(3,3) = 3 + 1 = 4
  Coefficients of 4th term in (x+Y)^4 = C(4,4) = C(3,3) + C(3,0) = 3 + 1 = 1
  Example : Find coefficients in (x+y)^5 and in (x+y)^6
  
  Coeff in (x+y)^4 : 1, 04, 06, 04, 01
  Coeff in (x+y)^5 : 1, 05, 10, 10, 05, 1
  Coeff in (x+y)^6 : 1, 06, 15, 20, 15, 6, 1
  Example : Find coefficient of (x^5)*(x*4) in expansion of (x+y)^n.
  
  Find C(n,r)
  
  C(n,r) is the coefficient (x^(n-r))*(y^r) in expansion of (x+y)^n
  x power + y power = n.
  
  Find n and r
  
  n = 5 + 4 and r = 4.
  Coeficient of (x^5)*(y^4) = C(9,4)
  C(9,4) = 9*8*7*(9-4+1)/4!
  C(9,4) = (9*8*7*6)/(4*3*2*1)
  C(9,4) = 126.
  Go to Begin
  AL 07 03. Coefficients of binomial expansion and Pascal triangle
  
  [Defintion] Pascal triangle : It is coeff of binomial expansion
  
  1 .................0..... Coeff in (x+y)^0
  1, 01 ................... Coeff in (x+y)^1
  1, 02, 01 ............... coeff in (x+y)^2
  1, 03, 03, 1 ............ coeff in (x+y)^3
  1, 04, 06, 04, 1 ........ coeff in (x+y)^4
  1, 05, 10, 10, 5, 1 ..... coeff in (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
  
  AL 07 04. Sequence in Pascal triangle
  
  Sequence along r0
  
  r0 sequence : 1, 1, 1, 1, ....
  nth term :T(n) = 1
  Sum of n terms = S(n) = Sum[n] = n
  Sequence along r1
  
  r0 sequence : 1, 2, 3, 4, 5, ....
  nth term :T(n) = n
  Sum of n terms = S(n) = Sum[n] = n*(n+1)/2
  Sequence along r2
  
  r2 sequence : 1, 3, 6, 10, 15, 21, 28, 36, ....
  T(n) = nth term = n*(n+1)/2
  S(n) = Sum[n*(n+1)/2] = n*(n+1)*(n+2)/6
  Sum[C(n+1, 2)] = C(n+2, 3)
  This is known as numbers in triangular number pattern
  Properties of this pattern
  
  1st difference is r1 sequence
  2nd difference is r0 sequence with common difference d = 1
  Properties of sequence of numbers in triangular sequence
  
  [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
  [Example] Prove that Sum[C(n+1,2)] = C(n+2,3)
  
  Since C(n+1,2) = n*(n+1)/2!
  Since C(n+2,3) = n*(n+1)*(n+2)/3!
  Hence Sum[C(n+1,2)] = C(n+2,3)
  
  Go to Begin
  
  AL 07 05. What is r3 series ?
  
  Sequence
  
  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
  
  AL 07 06. What is the sequence along r4 column ?
  
  Sequences
  
  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
  
  AL 07 07. What is the sequence along r5 column ?
  
  [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]
  
  Sum[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
  
  AL 07 08. Find Sum[n^4] by using Sum[C(n+3,4)] = C(n+4,5)
  
  Method
  
  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
  
  AL 07 09. Fibonacci's sequence in Pascal triangle
  Text and diagram
  
  in Fibonacc's pattern
  Example
  
  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
  
  AL 07 10. Series from C(n,r)
  
  [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]
  
  Study subject | Sum[n^2] = n*(n+1)*(2*n+1)/6
  Study subject | Sum[n^3] = (n*(n+1)/2)^2
  Study subject | Binomial theorem in calculus
  Sum[n^4] = (6*n^5 + 5*n^4 + 10*n^3 - n)/30
  
  Go to Begin
  
  AL 07 11. Constant oefficient in (x + 1/(x^2))^n
  
  The rth term
  
  rth term = C(n,r)*(x^(n-r))*(1/(x^2))^r)
  = C(n,r)*(x^(n-3*r))
  Constant term
  
  Power n - 3*r = 0 and r is integer
  Hence r = n/3
  Hence n must be multiple of 3.
  
  Go to Begin
  
  AL 07 012. Expand Sqr(1+x^2) into series
  
  By binomial theory
  
  Sqr(1 + x^2)
  = (1 + x^2)^(1/2)
  = 1 +(1/2)*(x^2) +((1/2)*(1/2-1)/2!)*((x^2)^2) +((1/2)*(1-1/2)*(1/2-2)/3!)*(x^2)^3 + ...
  = 1 + (1/2)*(x^2) - (1/((2^2)*2!)*(x^4) + (1*3/(2^3)*3!)*(x^6) + ...
  Find Sqr(2) using above series
  
  Let x = 1
  Sqr(2) = 1 + 1/2 - 1/8 + 1/16 - (1*3*5)/((2^4)*4!) + (1*3*5*7)/((2^5)*5!) - ...
  = 1 + 0.5 - 0.125 + 0.0625 - 0.03906 + 0.02734 - 0.025078....
  = 1.40070
  Error = 1.41421 - 1.40070 = 0.01351 =0.95%
  What is the error if we use the 8th term ?
  
  Go to Begin
  
  AL 07 13. Use Pascal triangle write down coefficient of expansion terms of (x+y)^7
  
  Pascal triangle
  
  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
  
  AL 07 14. Expansion of (x+1)^n
  
  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.
  Expand 1/(1+x^2) into series
  
  1/(1+x^2)
  = (1 + x^2)^(-1)
  = 1 +(-1)*(x^2) +((-1)*(-1-1)/(2!))*((x^2)^2) +((-1)*(-1-1)*(-1-2)/(3!))*((x^2)^3)+ ...
  = 1 - x^2 + x^4 - x^6 + x^8 + .....
  
  Go to Begin
  
  AL 07 15. Find coefficient 0f (x^3)*(y^5) in expansion (x+y)^n
  
  [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
  
  AL 07 16. Expansion of 1/(1 + x^2)^2
  
  Expand 1/(1+x^2) into series
  
  1/(1+x^2)
  = (1 + x^2)^(-1)
  = 1 +(-1)*(x^2) +((-1)*(-1-1)/(2!))*((x^2)^2) +((-1)*(-1-1)*(-1-2)/(3!))*((x^2)^3)+ ...
  = 1 - x^2 + x^4 - x^6 + x^8 + .....
  
  Go to Begin
  
  AL 07 17. Quiz for binomial theorem
  
  Questions
  
  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
  
  AL 07 18. 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.
  
  n = 100
  Coefficient = C(100,99) = 100
  
  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
  
  AL 07 00. Outlines
  
  Binomial theory
  
  (x + y)^n = C(n,0)*x^n + C(n,1)*(x^(n-1))*(y)+ C(n,2)*(x^(n-2))*(y^2) + ....
  Coefficients of expansion C(n,r) for r = 0,1,2,3....n
  C(n,r) = n*(n-1)*(n-2)*...*(n-r+1)/r! = coefficient of (r+1)th term
  Series from C(n,r) : Summation 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)
  Summation 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
  
  Study subject | Sum[n^2] = n*(n+1)*(2*n+1)/6
  Study subject | Sum[n^3] = (n*(n+1)/2)^2
  Study subject | Binomial theorem in calculus
  Sum[n^4] = (6*n^5 + 5*n^4 + 10*n^3 - n)/30
  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
  
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