.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.