Number System

Formulas for Number System and Basic Concept

   1. (a + b)² = a² + b² + 2ab        

2. (a - b)² = a² + b² - 2ab

3. (a + b)² - (a - b)² = 4ab   

4. (a + b)² + (a - b)² = 2 (a² + b²)

5.  (a² - b²) = (a + b) (a - b)

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

7.  (a³ + b³) = (a +b) (a² - ab + b²)       

8.  (a³ - b³) = (a - b) (a² + ab + b²)

9.  (a³ + b³ + c³ -3abc) = (a + b + c) (a² + b² + c²-ab-bc-ca)

 If a + b + c = 0, then a³ + b³ + c³ = 3abc.

COMMON SERIES

Sum of first n natural no’s=n(n+1)/2

Sum of first n odd numbers= n²

Sum of first n even numbers= n(n+1)

Sum of square of first n natural no’s  =n(n+1)(2n+1)/6

Sum of cubes of first n natural  no’s   = [n(n+1)/2]².

Arithmetic Progression(A.P.)

The nth term of this A.P. is given by

  Tn =a+ (n - 1) d.

The sum of n terms of this A.P.

  Sn = n/2 [2a + (n - 1) d]

  Sn= n/2 [first term + last term]

Geometrical Progression (G.P)

The nth term of this G.P. is given by

  Tn = ar^n-1

The sum of n terms of this G.P.

  Sn=     a(1-rⁿ)      when r<1
               (1-r)

  Sn=     a(rⁿ-1)     when r>1
                (r-1)

Frequently asked questions:

Question 1:Find the sum of first 100 natural numbers.

Solution:

Let S be the required sum.

Therefore, S = 1 + 2 + 3 + 4 + 5 + .................... + 100

Clearly, it is an Arithmetic Progression whose first term = 1, last term = 100 and number of terms = 100.

Therefore, S =100/2 (100 + 1), [Using the formula S = n/2(a + l)]

                  = 50(101)

                  = 5050
Therefore, the sum of first 100 natural numbers is 5050.

Question 2:Find the sum of the squares of first 50 natural numbers.

Solution:

We know the sum of the squares of first n natural numbers (S) = n(n+1)(2n+1)/6

Here n = 50

Therefore, the sum of the squares of first 50 natural numbers = 50(50+1)(2×50+1)/6

= 50×51×101/6=  42925

Question 3: The sum of all two digit numbers divisible by 5 is

Solution:

Required numbers are 10,15,20,25,...,95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)x5=95,then n=18.
Required sum=n/2(a+l)=18/2(10+95)=945.

Question 4: (51 + 52 + 53 + .........+100) is equal to

Solution:

51 + 52 + 53 + ...........+ 100

 = (1 + 2 + 3 + .... + 100) - (1 + 2 + 3 + ...... + 50)

 = It is in the form of n(n+1)/2 series summationn(n+1)/2 series summation

= n1 = 100 , n2 = 50

=[100(100+1)/2] − [50(50+1)/2]

 =  (5050 - 1275) = 3775

Question 5: What is the sum of all two-digit numbers which leave remainder 5 when they are divided by 7?

Solution:

The two digit natural numbers which leave a remainder 5, when divided by 7 are
12, 19, 26 ...., 89, 96. 

∴ 12, 19, 26, ...., 89, 96 is an A.P. whose first term a = 12 and common difference d = 7. 

 Tn.  Then, Tn = a + (n – 1) d,

  ⇒ 96 = 12 + (n – 1) 7 ⇒ 84 = (n – 1) 7 ⇒ n – 1 = 12 ⇒ n = 13

∴ Required Sum = n/2 (a + l) = 13/2(12 + 96)  = 132132 × 108  =13 × 54 = 702.