.Find the sum  of series of even squares in natural series .

S²_e = 2² + 4² + 6² + 8² ……… (2n)²

        = 2² [ 1² + 2² + 3² ………. n² )   where n = N_H/2

     = 4 [ n(n+1)(2n+1)/6] = (2/3)n(n+1)(2n+1)

 Find th sum of series of odd squares in natural series

  S²_o = 1² + 3² + 5² + 7² ……… (2n – 1)²

The sum can be determined by finding out the mean common difference.

With two terms 1² + 3² = 10 =(2+d)   or  d = 8

With three terms  1² + 3² + 5² = 35 = 3 +3d or d = 10 + 2/3  = 8 + (2 +2/3)

With four terms 1² + 3² + 5² + 7² = 84 = 4 +6d or d = 13 +1/3 = 8 + 2 ( 2 + 2/3)

With n terms 1² + 3² + 5² + 7² ……… (2n – 1)²   , n + n(n-1)<d>/2  where the mean <d> = 8 +(n-2)[2 +2/3] = 8 + 8n/3 – 16/3 = (8/3) (n+1)

S²_o = n [ 1 + (n-1) <d>/2]

     = n [ 1 + (n-1)/2 [8/3 (n+1)]

     = n [ 1 +(4/3)(n² -1)] = (n/3) [4n² – 1]

 

 

 Find the sum of the given series  (1x2) + (2x3) + (3x4)……… [(n-1)n]  

       1²  =  1

      2²  =  2 + 1.2

    3²  =  3  + 2.3

    4²  = 4 + 3 x 4

    n² = n + (n-1)n

Summing up all the equations, ( 1² + 2² + 3² + 4² ……. n²) =( 1 + 2 + 3 + 4 ……. n)+ [(1x2) + (2x3) + (3x4)……… [(n-1)n]

We know , ( 1² + 2² + 3² + 4² ……. n²) = n(n+1)(2n+1)/6

( 1 + 2 + 3 + 4 ……. n) = n(n+1)/2

Hence , [(1x2) + (2x3) + (3x4)……… [(n-1)n] = n(n+1)(2n+1)/6  - n(n+1)/2

On simplification it reduces to n(n²-1) /3

 

 

The sum of series of successive cubes is equal to square of the sum of root numbers.

1³ + 2³ = 1+8 = 9 = 3² = (1+2)²

1³ + 2³  + 3³ = 1+8 +27 = 36 =6² = (1+2 + 3)²

1³ + 2³  + 3³ + 4³= 1+8 +27 + 64 = 100 =10² = (1+2 + 3 + 4)²

1³ + 2³  + 3³ + 4³ …… n³ =  (1+2 + 3 + 4…… n )² = [n(n+1)/2]²

This can be proved b considering this series as an arithmetic series with varying common difference.

If there are two terms 1³ + 2³  = 9 = 1 +(1+d₁ ) = 1 + (1+d) or d = d₁ = 7

With three terms 1³ + 2³  + 3³ = 36 = 3 + d₁ + d₂ = 3 + (1+d) +(1+2d) = 3 + 3d

                               d₁ + d₂ = 33 ; d₂ = 26 and d = 11 = 7 +4

With four terms  1³ + 2³  + 3³ + 4³ = 100 = 4+ d₁ + d₂ + d₃ = 4 + 6d

                                d₁ + d₂ + d₃  = 96 , d₃ = 63 and d = 16 = 7 + 4 +5

The mean d varies as d = 7+  [(n+1)(n+2)/2 – 3x4/2] = 7 + [ (n² +3n +2)/2 -6] = 1 +(n² +3n +2)/2 = (n² +3n +4)/2 where n ≥ 2

S³ = 1³ + 2³  + 3³ + 4³ …… (n-1)³  where  <d> =  (n² +3n +4)/2

 

   = n [ 1 + (n-1)/2 d] = n { 1 + [(n-1)/2][ (n² +3n +4)/2] = [n(n+1)/2]²

Sum of successive square numbers

S = 1² + 2² + 3² + 4² +……….   + n² , where n is the order of the number in the series which is equal to its value

These squares are grouped into series of odd squares 1² + 3² + 5² +……….   + (2n-1)²

And series of even squares 2² + 4² + 6² +……….   + (2n)²

Where n is the order of the odd/even number in the series with values  N_o – (2n -1) and N_e= 2n. It is noted that the sum of squares of all odd numbers from 1 to (2n-1) in natural series is 1/6 times the product of three successive numbers (2n-1)2n(2n+1)

 1² = (1x2x3)/6 = 1

1² +3² = (3x4x5)/6 = 10

1² +3² + 5² = (5x6x7)/6 =35 and so on.

1² + 3² + 5² +……….   + (2n-1)² = (2n-1)2n(2n+1)/6= (n/3)(4n² – 1)

In terms of the odd numbers  N_oH = 2n -1, 2n = (N_oH + 1) and 2n+1 = N_oH +2. In terms of N_oH ,the sum becomes So² = N_oH (N_oH +1) (N_oH +2)/ 6

 

Like wise the sum of all even numbers from2 to 2n in the natural series is 1/6 times the product of three successive numbers 2n (2n+1) (2n+2). For example,

 

2²  = (2x3x4)/6  = 4

2² +4² = (4x5x 6)/6 = 20

2² +4² + 6² = (6x7x8)/6 = 56

In general 2² + 4² + 6² +……….   + (2n)²  = 2n (2n+1)(2n+2)/6 = (2/3) n (n+1)(2n+1)

In terms of the even numbers  N_eH = 2n , n = (N_eH /2) , (n+1) = (N_eH +2)/2 and 2n+ 1 = N_eH +1. In terms of N_eH, the sum becomes S_e² =(1/6) N_{eH (}N_eH+1) ₍N_eH +2)

The sum  all squares in natural series is the sum of the sum of odd squares and sum of even squares S² = S_o² + S_e²

In terms of order the numbers \ (n/3) (4n² -1) + (2n/3)(n+1)(2n+1) = (n/3)(2n+1)(4n+1) . In terms of the actual value of the number N_H = 2n

S = N_H (N_H +1)( 2N_H +1)/6

 

There is yet another way to derive the general expression for the sum of squares in natural series S = 1² + 2² + 3² + 4² + ………… This can be considered as an arithmetic series with different common difference . By estimating the mean common difference , we can use the same formula used for arithmetic series with a common difference. The common difference may be different, but it varies gradually ,which can be estimated.

 

1² + 2² = 5 = 1 +(1+d₁) = 1 + (1+d) or d= d₁ = 3

1² + 2² + 3²  = 14 = 3+ d₁ + d₂ = 3 + 3d; d₂ + d₂ = 11 and d₂ = 8 and d = 11/3 = 3 +(2/3)

1² + 2² + 3²  + 4² = 30 = 4 + d₁ + d₂  + d₃ = 4 + 6d ; d₁ + d₂  + d₃ = 26  and d₃ = 15. d = 13/3 = 3+(2/3) +(2/3) = 4 + (1/3)

1² + 2² + 3²  + 4² + 5² =55 = 5 + d₁ + d₂  + d₃ + d₄ = 5 +10d ; d₁ + d₂  + d₃ + d₄ = 50 and d₄ =24 , d = 4 + (1/3) + (2/3) = 5.

It is noted that the mean d is gradually increasing  3 + (2/3) (n-2)

If there are n terms  1 +(1+d) + (1+2d) + (1+3d) …….. [1 +(n-1)d]

S = n + d[ 1+2+3……(n-1)] = n + n(n-1)d/2 = n[ 1 + (n-1)d/2]

Substituting the mean value of d, we get,

S = n { 1+ [(n-1)/2][ 3 + 2/3(n-2)]} = n { 1 + [(n-1)/2](9+2n-4)/3 } = n [1 +(n-1/2)(2n+5/3)] = n[(2n² +3n +1)/6 = n(n+1)(2n+1)/6   

  

 

1.solve a² + b²  = c² with the help of algebra where the factors of one of the square number is kown 

 With single variable a = α²

a²  = c^{2  -} b²  

α⁴ = (c-b)(c+b) .If c - b = α and c + b = α³ , which give c = α (α² +1)/2 and b = α (α^{2  -} 1)/2 . The solution then becomes (2α²)² +  [ α (α^{2  -} 1) ]² = [ α (α² +1) ]²

with two variables a = αβ      

α² β²  = (c-b) (c+b) . If c-b = α  and c+b = αβ² .By solving for b and c, we get a solution as

(2αβ)² + [α(β² – 1)]²  = [α(β² + 1)]²

The distribution of prime factors of a  to (c-b) and (c+b) may have different values ,however its product must be equal to the square of the prime factors of a. If If c-b = α²  and c+b = β² .By solving for b and c, we get a solution as

(αβ)² + (β² – α²⁾²  =  (β² + α²⁾²

(2) Solve a³ + b³ = c³ by using prime factor method

With single variable . let a = α^2. Then α^{6 =} (c – b) [(c-b)² + 3cb]. Under one possible distribution of prime factors (c-b) = α and [(c-b)² + 3cb].=  α² + 3cb = α⁵ or 3cb = α² (α³ -1) or b[= [α² (α³ -1)]/3c .Substituting this value in c-b ,  c -  [α² (α³ -1)]/3c = α  or  3c² – 3cα -[α² (α³ -1)] = 0. Solving this quadratic equation for c, we get

c =[3α ± √ (9α² + 12 α^5- 12 α² ) ]/6 = α/2 + (1/6) √ (12 α^3- 3 5² ) ] and b = c – α = - α/2 + (1/6) √ (12 α^{5 -} 3 α² ) ]. It gives its solution as (α ²)³ + [- α/2 + (1/6) √ (12 α^3- 3 α²)]³ = [ α/2 + (1/6) √ (12 α^5- 3 α²)]³ . Whatever may be the prime factors of a and its distribution among b and c, ,there will be integral solution. Ie., all a,b,c cannot have positive integers. This is called Fermat’s Last theorem.

With two varibales. Let a = αβ. Then  α³ β³ =   (c – b) [(c-b)² + 3cb].. If c-b = α² ,  then   [(c-b)² + 3cb]. = α⁴  + 3cb =  α β³ or b = [α (β³ – α³ )/3c. Substituting this value in c-b , c - [α (β³ – α³ )/3c.]^- = α²  or  3c² – 3c α² - α (β³ – α³ ) = 0. The solution of this quadratic equation becomes c = [3 α^{2 ±} √ 9 α⁴ + 12 α (β³ – α³ )]/6 . It is reduced to c = (α²/2) + (1/6) √  12 α β³ –3 α⁴ )  and b = c- α^{2 = -} (α²/2) + (1/6) √12 α β³ –3 α⁴ ) ^-.It gives the solution as (αβ)³ + [ (α²/2) + (1/6) √12 α β³ –3 α⁴)]³ = [(α²/2) + (1/6) √12 α β³ –3 α⁴)]³

(3) The prime factor method is application to solve a^x + b^y = c^z.,where the exponents are not same..S=olve a² + b³ = c^{3 .}for the integral values of a,b and c.

Let  a = =   α², then α⁴ = (c-b) [ (c-b)² + 3cb]. One possible distribution  is c-b = α and  (c-b)² + 3cb] = α² + 3cb = α³  or b = [α²  (α -1)]/3c  .Substituting this value of b in c-b  = α we get a quadratic equation 3c² – 3cα - α² (α -1) = 0, Solving for c, we get its two roots c =[(α/2) + (1/6)√ (12 α³ - 3 α² )]  and b = [ - α/2) + (1/6)√ (12 α³ - 3 α² )]]. The general form of the relation becomes ( α² )² + [(- α/2) + (1/6)√ (12 α³ - 3 α)]³ = [ (α/2) + (1/6)√ (12 α³ - 3 α)] . It is seen that when α = 1,7,19,37……. The root numbers become whole integers. When α = 7 ;  49² + 7³ = 14³

 (4) Solve a³  + b² = c² for integral values of a,b,and c.

 a³   = c²  - b²= (c-b) (c+b)

With single variable  a,  c-b =a and c+b = a² which give the value for b and c in terms of a.

c = a(a+1)/2 and b = a(a-1)/2. The relation in terms of a single variable a is a³ + [a(a-1)/2]² = [a(a+1)/2]²

 a=2 ; 2³ + 1²= 3²

a=3 ; 3³ + 3² = 6²

with two variables a = αβ; (αβ)³ = (c-b)(c+b) . One possible distribution of the prime factors is  c-b = α² β  and c+b = αβ^{2 .} It gives c = αβ ( β + α) / 2 and b = αβ (β - α) / 2. In terms of these variables, the relation becomes  (αβ)³ + [αβ (β - α) / 2]² = [αβ (β + α) / 2]²

α = 2,β = 3;  6³ + 3² = 15²

α = 2,β = 5; 10³ + 15² = 35²

One may show an addition of two three digits numbers where all the digits from 1 to 9 are used once only.

                                            n₁     n₂         n₃

                                           n₄      n₅        n₆

                                      ---------------------------

                                           n₇      n₈        n₀     

(100 n₁ + 10 n₂+ 9) + (100 n₄  + 10 n₅ + n₆ )  = 100 n₇ + 10 n_{8 +} n₉

The digital root of the sum of the digital roots of adders  D (n₁ + n_{2 +}  n_{3 +}  n₄ + n₅ + n₆ ) must be equal to the digital root of the additive sum D( n_{4 +} n₅ + n₆)  This is possible only if the digital roots of both sides are equal to 9.

Few solutions are

182 + 763 = 945

492 +183 = 675

592 + 146 = 738

782 + 154 = 936

193 + 275 = 468

159 + 327 = 486

349 + 218 = 567

529 + 317 = 846

There are very few solutions where the sum is a square.

143 + 586 = 729 = 27 x27 = 27²

194 + 382 = 576 = 24 x 24 = 24²

369 + 415 = 784 = 28 x 28 = 28²

. The sum of three two digits numbers give a three digits number. Get solutions in the summation all the digits from 1 to 9 are used once only.

(10 n₁ + n₂ ) + ( 10 n₃ + n₄ ) +(10 n₅ +n₆ ) = (100 n₇ + 10 n₈ + n₉ )

The digital roots must be same for both sides of the relation.D (n₁+n_{2 +} n₃ + n_{2 +}  n₄ + n₅ + n₆)                                 = D (n₇+n₈ + n₉)

A typical solution is 83 +54 +79 = 216 = 6³. By interchanging the digits of unit and 10 th decimal place  we get additional solution 84 +59 + 73 = 89 +53 + 74 = 216

  xyz + xyz + xyz = zzz  or xyz + xyz + xyz = yyy  or xyz + xyz + xyz = www, where w,x,y,z

take a digit from 1 to 9.. Find allowed values of w,x,y.z.

xyz + xyz + xyz = zzz

3 ( 100 x + 10 y+ z) = 100 z + 10 z + z = 111 z

      100x +10y + z = 37 z -à  100x +10y = 36 z-à  -à 50 x + 5 y = 18 z

One possible solution is z =5 , x= 1 and y =8  )185, which gives 185 x 3 = 555

xyz + xyz + xyz = yyy

3 ( 100 x + 10 y+ z) = 100 y + 10 y + y = 111 y

     100 x + 10y + z = 37 y --à 100x + z = 27 y

One possible  solution is  y =4 ,x=1 , z= 8 which gives 148 x 3 = 444

( x≠w, as it gives forbidden solutions)

If the identical digit in th sum is different from the digits of the adder

3(100x + 10y +z) = 100 w + 10w + w = 111 w

  (100x + 10y +z) = 37 w

 37 x 3n gives all identical digits.  Two solutions are available\

259 x 3 = 777 and 296 x 3 = 888

  

 Problem solving skill

(1)You are given odd numbers 1,3,5,7 and 9. Can you get 2,4,6 and 8  with any three same or different numbers using addition and subtraction only.?

It is not possible. Sum of any three odd numbers will always be odd., can never be even.

(2)  x⁵ + x⁴ + x³ + x² + x + 1= 0. Find the value of x.

The given expression can be written as x (x⁴ + x³ + x² + x + 1) = -1.There are two possibilities.

x = 1 and (x⁴ + x³ + x² + x + 1) = -1

x = -1 and (x⁴ + x³ + x² + x + 1) = 1

The former solution is not valid as (x⁴ + x³ + x² + x + 1) ≠ -1 if x =1.x = -1, satisfy both the condition.  

  Learning Mathematics through fun -Problem solving 

(1) 1  + 3 + 5 = 9 = 3 x 3 = 3²

Identify similar set of successive odd numbers whose sum gives a square.

(2n-1) + (2n+1) + (2n+3) = 6n +3 = 3(2n+1). To satisfy the condition, (2n+1) must be odd which must be a multiple of 3 and a square number.i.e., (2n+1) = 3,12,27,48,75 …….

It gives the required solution    10 +12 +14 = 36 = 6 x 6 = 6²    

                                              25 +27 + 29 = 81 = 9 x 9 = 9²   

                                               73 + 75 + 77 = 225 = 15 x 15 =  15²  

(2)  1+3+5+7 = 16 = 4 x4 = 4²

       Find out few more set of successive odd numbers whose sum gives a square

     (2n-1) + (2n+1) + (2n+3) + (2n+5) = 8(n+1)= 4 x 2(n+1)

     2(n+1) must be a square number  It is possible when n = 1,7,17,31 …….

13 +15 +17 +19 = 64 = 8x 8 = 8²   

33 + 35 + 37 + 39 = 144 = 12 x 12 = 12² 

61 + 63 + 65 + 67 = 256 = 16 x16 = 16²

(3) 3 + 5 = 8 = 2 x 2 x 2 = 2³  

Here the sum of two successive odd numbers give a cube. Can you spot out few such pairs.?

(2n-1) + (2n+1) = 4n ,To be a cube n = 2x³  or n = 2,16,54 ………. (31,33) ,(107.109) are two such apirs whose sum give cube.

31 + 33 = 64 = 4x4x4 = 4³  

107 + 109 = 216 = 6x 6 x6 = 6³

(4) What are the three successive odd numbers whose sum gives a cube ?

 (2n-1) + (2n+1) + (2n+3) = 6n +3 = 3 (2n+1). To give odd cube (2n+1) must be odd  or (2n+1) = .  9 , ,243 ------9 (2x-1)³  . It gives

7+9+11=27 = 3x3x3 = 3³  

241 + 243 + 245 = 729 = 9x9x9 = 9³

 Knowing the Algebraic formulae

Algebra is a branch of mathematics in which all unknown quantities are expressed as abstract symbols rather than specific numbers. Algebra is the branch of mathematics that helps in the representation of problems or situations in the form of mathematical expressions. By analyzing the mathematical expression one can understand various dependency of various factors associated with the problem.  Algebra is helpful to develop skill in problem solving . There ar many Algebraic formulae  Let us try to learn these algebraic formulae from fundamental concepts in mathematics.

a² – b² = (a + b) (a - b)

Adding  and subtracting a common  term ab with a² – b^2,  we get

a² + ab – ab - b² where a is commen in the first two terms andb is common in the last two terms.

a(a+b) – b(a+b), where (a+b) is common in both the terms, It is simplified into (a+b)    (a-b)

(a + b)² = a² + 2ab + b²

(a+b)²  = (a+b) (a+b) = a²  + ab + ab + b^2, = a²  + 2ab + b^2,. It gives a supplementary formula  a² + b² = (a + b)² – 2ab.

(a – b)² = a² – 2ab + b²

(a-b)² = (a-b)(a-b) = a²  - ab - ab + b^2, = a²  - 2ab + b^2. It gives a supplementary formula  a² + b² = (a - b)² + 2ab . It shows that (a-b)² , (a² + b²) , (a+b)²  are always in arithmetic progression with a common difference 2ab.

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

(a+b+c) (a+b+c) = a²  ab + ac + ab + b² + bc + ac + bc + c² =  a² + b² + c² + 2ab + 2bc + 2ca.

(a – b – c)² = (a-b-c) (a-b-c) =  a² + b² + c² – 2ab + 2bc – 2ca.

(a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab (a + b)

(a+b)³  = (a+b) (a+b) (a+b) = (a² +2ab + b²⁾)(a+b) = (a³ + 2a² b +ab² ) + (a²b + 2ab² + b³) = a³ + 3a² b + 3ab² + b³ . The middle two terms have a common factor 3ab which reduces the formula into (a + b)³  =  a³ + b³ + 3ab(a + b).It provides another supplementary formula a³ + b³  =(a + b)³  - 3ab(a+b). Since (a+b) is common it is changed into  a³ + b³  =(a + b)³  - 3ab(a+b) = (a+b) [(a+b)² – 3ab] =(a+b)(a² –ab + b² ]

The same formula can be obtained directly from a³ + b³  by imply adding and subtracting  common and conjugate pairs a²b + ab²

a³ + b³ = a³ + a²b + b³ + b²a - a²b - ab²

The first two terms have a common factor a², the middle two terms have a common factor b², while the last two terms have a common factor ab..It reduces into

a²(a+b) + b² (a+b) – ab(a+b) = (a+b) (a² –ab + b² )

(a – b)³ = a³ – 3a²b + 3ab² – b³ = a³ – b³ – 3ab (a – b)

(a-b)³  = (a-b) (a-b) (a-b) = (a² -2ab + b²⁾)(a-b) = (a³ - 2a² b +ab² ) + (-a²b + 2ab² - b³) = a³ - 3a² b + 3ab² - b³ . The middle two terms have a common factor 3ab which reduces the formula into (a - b)³  =  a³ - b³ - 3ab(a - b).It provides another supplementary formula a³ - b³  =(a - b)³ + 3ab(a-b). Since (a-b) is common it is changed into  a³ -  b³  =(a - b)³  -+ 3ab(a- b) = (a-b) [(a-b)² + 3ab] =(a-b)(a² + ab + b² ]

The same formula can be obtained directly from a³ - b³  by imply adding and subtracting  common and conjugate pairs a²b + ab²

a³ - b³ = a³ - a²b - b³ + b²a + a²b - ab²

The first two terms have a common factor a², the middle two terms have a common factor b², while the last two terms have a common factor ab..It reduces into

a²(a-b) + b² (a-b) –+ab(a-b) = (a-b) (a² + ab + b² )

By following  a similar procedure one can establish algebraic formula for a^{n –}  bⁿ   where the exponent may have any integral value.. For example at first let us consider a⁴ – b^{4 .}.In the first step add and subtract  ( a³ b – ab³) ,a conjugate pair with power 4

a⁴ – b⁴  = (a+b) ( a³ – a²b + ab² – b³ )                                                                                             a⁴ – b⁴  = (a - b) ( a³ + a²b + ab² + b³ )

a⁴ – b⁴  = a⁴ – b⁴   + ( a³ b – ab³) - ( a³ b – ab³)

            = a³ (a+b) – b³ (a+b)  - ( a³ b – ab³)  

In the second step add and subtract a² b²  a conjugate pair  with power 4

a⁴ – b⁴  = a³ (a+b) – b³ (a+b)  - ( a³ b – ab³) - a² b²  + a² b²

           = a³ (a+b) – b³ (a+b) – a²b (a+ b) + ab² (a+b)

(a+b) is common in all the terms which reduces the relation

       a⁴ – b⁴  = (a+b) ( a³ – a²b + ab² – b³ )

a⁴ – b⁴   can be expressed in another way.In the first step subtract and add (-a³ b + ab³), the conjugate pair with power 4.

a⁴ – b⁴  = a⁴ – b⁴   + ( -a³ b + ab³) - (- a³ b + ab³)

            = a³ (a-b) + b³ (a-b)  + ( a³ b – ab³)

In the second step add and subtract a² b²  a conjugate pair  with power 4

a⁴ – b⁴  = a³ (a-b) + b³ (a-b)  + ( a³ b – ab³) + a² b²  - a² b²

           = a³ (a-b) – b³ (a-b) +a²b (a- b) + ab² (a-b)

(a - b) is common in all the terms which reduces the relation

       a⁴ – b⁴  = (a - b) ( a³ + a²b + ab² + b³ )

This can be obtained directly  by factorizing the therm a⁴ – b⁴

 a ⁴ – b⁴ = (a² + b² )(a² - b²) = (a² + b² )(a+b) (a-b)

a⁴ – b⁴ =  (a+b) [(a-b) (a² + b² )] = (a+b)( a³ –a²b + ab² – b³ )

a⁴ – b⁴ =  (a-b) [(a+b) (a² + b² )] = (a-b)( a³ +a²b + ab² + b³ )

In the same way one can prove

a⁵ – b⁵ = (a-b) (a⁴ + a³b + a²b² + ab³ + b⁴)

a⁶  –  b⁶ = (a-b) (a⁵ + a⁴b + a³b² + a²b³ + ab⁴ + b⁵)

If n is a natural number aⁿ– bⁿ = (a-b) (a^n-1 + a^n-2b + a^n-3b² + …… ab^n-2+  b^n-1)

aⁿ – bⁿ is divisible by (a-b) what ever may be the value of the exponent. If the exponent is even it is divisible by both (a-b) and (a+b).

aⁿ – bⁿ  can also be expressed as a product with (a+b) for odd exponents. For example let us consider the case a⁵ – b^5.. In the first stage add and subtract a conjugate pair  for 5 th power a⁵ – b⁵   = a⁵ – b⁵  + a⁴b – b⁴a  + ( - a⁴b + b⁴a) =a⁴(a+b) – b⁴ (a+b) – a⁴b + b⁴a. in the second stage add and subtract –a³ b² + a² b³    we get a⁵ – b⁵   = (a+b) (a⁴ – b⁴) – a³b (a+b) + ab³ (a+b)  + a³ b² – a² b³ = (a+b) (a⁴ – a³b +  a³b -  b⁴)  + a² b² (a-b)

a³ – b³ = (a+b) { a² – b²) – ab(a-b)

a⁵ – b⁵   = (a+b) (a⁴ – a³b +  a³b -  b⁴)  + a² b² (a-b)

a⁷ – b^{7 =} (a+b)( a⁶ – a⁵b + a⁴b^2- a²b⁴ +a b⁵ – b⁶) – a³b³(a+b)

In general

a²ⁿ⁺¹ – b^{2n +1} = (a+b) [a²ⁿ – a^2n-1b + a^2n-2 b²

                                    -b²ⁿ + b^2n-1a – b^2n-2 a²]  + (-1)ⁿ aⁿbⁿ (a-b)

if n is even (n = 2k), aⁿ + bⁿ = (a + b)(a^n-1 – a^n-2b +…+ b^n-2a – b^n-1)

If n is odd (n = 2k + 1), aⁿ + bⁿ = (a + b)(a^n-1 – a^n-2b +a^n-3b²…- b^n-2a + b^n-1)

(a + b + c + d)² = a² + b² + c² + d² + 2ab + 2 ac + 2ad+2bc + 2bd + 2cd

(a + b + c + d)² = [ (a+b) + (c+d)]² = (a+b)²  + (c+d)² + 2(a+b)(c+d)

                                                        = a² + b² + 2ab + c² + d² + 2cd + 2ac + 2ad + 2bc + 2bd

                                                       = a² + b² + c² + d² + 2ab + 2ac + 2ad + +2bc + 2bd +2cd

Solution of Quadratic equation

The general form of  a quadratic equation is ax² + bx + c = 0 ,where a,b,c are some coefficients, invariables, while x has certain specific values which satisfy the equation, called solutions or roots.

Determination of roots of the quadratic equation

Step-1 : Divide by a, the coefficient of x^{2 ;}   x² + (b/a) x = - c/a

Step-2 : Make the terms in LHS to be a square .For that  the coefficient of x = b/a is halved and then squared .It gives b²/4a². And is added on both sides

     x² + (b/a) x +  b²/4a²= - c/a + b²/4a²

       (x + b/2a)² =  b²/4a²  - c/a

       (x + b/2a)  =  ±[ (b/2a) ²  - c/a]^1/2

       x = [ -b  ± (b² – 4ac)^1/2 ]/2a

Sum of two roots

Let x₁ and x₂ be two roots that satisfy the   quadratic equation  ax² + bx = -c

 ax₁ ² + bx₁ = -c and   ax₂ ² + bx₂ = -c

Subtracting one from the other

a( x₂² – x₁²) + b(x₂ – x₁) = (x₂ – x₁) [ a(x₂ + x₁) + b] = 0

If the two roots are different  a(x₂ + x₁) + b = 0 or x₁ + x_{2  = -} b/a

Product of two roots

ax₁² + bx₁ = - c and   ax₂² + bx₂ = - c Adding together we

a( x₂² + x₁²) + b(x₂ + x₁) = -2c_.

 Adding 2x₁ x₂  both side and reduce the equation  to

a( x₂² + x₁²)² + b(x₂ + x₁) = -2c_. +  2x₁ x₂

Substitute the value for sum of the two roots x₂ + x₁ = -b/a we have x₁x₂  = c/a