BLOGGS ON FILMS , DAY TODAY EVENTS ,FESTIVALS , FASHION, LEGAL ACT, TECHNOLOGY, SOCIAL ISSUES , INTERESTING STORIES ETC.
Tuesday, 24 November 2020
Line & It's Equations
Sunday, 15 November 2020
LAPTOP (EK MAUT KI KATHA)
" LAPTOP " (EK MAUT KI KATHA)
Monday, 9 November 2020
PASCAL TRIANGLE & FIBONACCI NUMBERS
PASCAL TRIANGLE
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
PASCAL TRIANGLE GIVES THE BINOMIAL
COEFFICIENT IN THE EXPANSION OF BINOMIAL THEOREM.
READ THE BLOG ON "INDIAN RAILWAY "
FOR Example
1) In the expansion of (a+b)¹ = a + b = 1a+ 1b
The coefficient of a & b is 1 & 1 .
The Coefficient of 1st term & 2 nd term is 1 & 1 i.e. 1 1
This is shown in the second row of Pascal's Triangle .
2) In the expansion of (a+b)² = a² +2ab+b² =1a² +2ab+1b²
The Coefficient of 1st term is 1 , Coefficient of 2nd term is 2 & Coefficient of 3rd term is 1 i.e. 1 2 1
This is shown in the 3rd row of Pascal's Triangle.
3)In the expansion of (a+b)³ = a³ +3a²b+3ab²+b³
= 1a³ +3a²b+3ab²+1b³
The Coefficient of 1st term is 1, coefficient of 2nd term is 3, coefficient of 3rd term is 3 & Coefficient of last term is 1. i.e.
The Coefficient are 1 3 3 1
This is shown in the 4th row of Pascal's Triangle.
4) Same way expansion of of (a+b)⁴
(a+b)⁴= a⁴ +a³b¹+a²b²+a¹b³+b⁴=
= 1a⁴+4a³b¹+6a²b²+4a¹b³+1b⁴
The Coefficients are 1 4 6 4 1
This is shown in the 5th row of Pascal's Triangle .
This way Next row in Pascal triangle shows
the Coefficients of (a+b)⁵ i.e. 1 5 10 10 5 1 & so on .
NOTE : 1) IN THE THIRD ROW OF PASCAL TRIANGLE 2
IS OBTAIN BY ADDING NUMBER IN PREVIOS ROW i.e. 1+1
2)Fourth row is obtain from 3rd row. i.e. 3 is obtain from 1+2 & next 3 is obtain from 2+1.
3)simillarly fifth row is obtain from 4th row i.e. 4 is obtain from1+3 , 6is obtain from 3+3 , next 4is obtain from 3+1 & so on.
THE FIBONACCI SEQUENCE IS ALSO OBTAIN FROM
PASCAL'S TRIANGLE.
The Sequence is 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1497 ....... so on
READ THE BLOG ON "BINOMIAL THEOREM"
In Fibonacci Sequence 1st & 2nd term is 1 i.e. sequence is 1 1
3rd term is 1+1 =2 i.e now the sequence is 1 1 2
4th term is 1+2=3 i.e now the sequence is 1 1 2 3
5th term is 2+3=5 i.e now the sequence is 1 1 2 3 5
6th term is 3+5=8 i.e. now the sequence is 1 1 2 3 5 8
7thet erm is 5+8=13 i.e. now the sequence is 1 1 2 3 5 8 13
8the term is 8+13=21 , 9th term 13+21=34 , 10 th term 21+34=55 & so on .
Sunday, 8 November 2020
BASIC ALJEBRA FORMUAE
BASIC ALJEBRA FORMUAE
1) (a+b)¹ =a +b
2) (a+b)² =a² +2ab +b²
3) (a-b)² =a² -2ab +b²
4) (a+b)³ =a³ +3a²b+3ab²+b³
5)(a-b)³ =a³-3a²b+3ab²-b³
6)(a+b)⁴=a⁴+4a³b¹ +6a²b² +4a¹b³ +b⁴
7)(a+b)⁵=a⁵+5a⁴b¹+10a³b²+10a²b³+5a¹b⁴+b⁵
8) (a²-b²) =(a+b) (a-b)
9) (a+b)(a-b) =a²-b²
10) (a³+b³) =(a+b) (a²-ab+b² )
11) (a³-b³) = (a-b) (a²+ab+b² )
12)(a+b+c)² =a²+b²+c²+2ab+2bc+2ca
13) (a+1/a)² = a²+2+1/a²
.i.e a²+1/a² = (a+1/a)²-2
14) (a-1/a)² =a²-2+1/a²
i.e. a²+1/a² = (a-1/a)² +2
15) (a+1/a)³ = a³+3a+3.1/a+1/a³
16) (a+b)²-2ab =a²+b²
17) (a+b)² - 4ab = (a-b)²
18) (a-b)² +4ab= (a+b)²
19) (x +a) (x+b) =x (x+b) +a (x+b)
= x² +x.b +a.x+a.b
20) (x +a) (x-b) = x (x-b) +a (x-b)
=x² - x.b + a.x - a.b
21) (x-a) (x+b) = x (x+b) - a(x+b)
=x² +x.b - a.x - a.b
22) (x-a) (x-b) = x (x-b) -a (x-b)
= x² - x.b - a.x + a.b
READ THE BLOG ON ANGLE & IT'S TYPE
Laws of Indices :
1) (a.b)ⁿ=aⁿ.bⁿ (Power is common base is different)
2) aⁿ. aⁱ= aⁿ⁺ⁱ (Base is same power is different)
3) 1/aⁱ = a⁻ⁱ
3) aⁿ/aⁱ =aⁿ➗ aⁱ = aª⁻ⁱ
4) (aⁿ)ⁱ = aⁿⁱ
5)( (aⁿ)ⁱ )ˡ = (a)ⁿⁱˡ
6) aº = 1
7) a¹ = a
8) a⁻ⁿ = 1/ aⁿ
9) (a)¹/ⁿ = n√a e.g. ∜a = (a)¹/4
10) (a)ⁿ/ˡ =1 / (a)ˡ/n
11) (a/b)ⁿ = aⁿ/bⁿ
READ THE BLOG ON " INDIAN RAILWAY "
Note :
1) (-) × (-) = +
2) (-) × (+) = -
3) (+) × (-) = -
4) (+) × (+) = +
Saturday, 7 November 2020
BINOMIAL THEOREM
Binomial Theorem
` Factorial Notation :For a natural number n , the factorial
of n is written as n! or ∟n & read as " n factorial " is the
productof n natural numbers from 1 to n .
i.e. n! =n (n-1)(n-2)(n-3)(n-4) ........4 . 3 . 2 . 1
=n (n-1)!
=n.(n-1).(n-2)!
e.g.
0!=1
1!=1
2!=2✕1=2.1=2
3!=3✕2✕1=3.2.1=6
4!=4✕3✕2✕1=4.3.2.1=24
5! =5✕4✕3✕2✕1 =5.4.3.2.1=120
6!=6✕5✕4✕3✕2✕1=6.5.4.3.2.1=720
7! =7✕6✕5✕4✕3✕2✕1=7.6.5.4.3.2.1=5040
......... so on.
Read the Blog on ANGLE & IT'S TYPE
COMBINATION : A Combination of a set of n object taken r
at a time without repetition is known as Combination in Mathematics
i.e. Out of n object if r object is selected at time ,
where order is not important. It is denoted by
nCr or C(n ,r) or nCr .
nCr = n! = n!/r!(n-r)!
r! (n-r)!
e.g. 10C4 =10!/ 4! (10-4)!
= 10.9.8.7.6!
4.3.2.1. 6!
= 10.9.8.7 = 210
4.3.2.1
The Above calculation is according to the defination
of Combination.
The Calculation can also be done like this .
10C4 = 10.9.8.7
4.3.2.1
20C6 = 20.19.18.17.16.15
6.5.4.3.2.1
i.e. If the value of r is 4 then just write
Product of n to (n-3) narural number in the numerator
and Product of 4 natural number from1to 4 in denominator
as in 1st example.
Note : nC0=nCn=1 & nC1=n
Bionomial Theorem :
(a+b)¹ = a + b
(a+b)²= a² +2ab+ b²
(a+b)³= a³ + 3a²b + 3ab² + b³
(a+b)⁴= ?
When the higher powers of (a+b) comes then it
becomes difficult for us to remember. If someone
asks us what is 29th power of (a+b) =?
Then it becomes difficult for us to answer .
At that time we can use Binomial Theorem
to Find Higher Powers of (a+b) or any variable.
READ THE BLOG ON PHYSICS & PHYSICAL QUANTITIES
(a+b)ⁿ =nC0aⁿbº +nC1aⁿ⁻¹b¹ +nC2aⁿ⁻²b²+nC3aⁿ⁻³b³ +........+nCnaºbⁿ
For Example :
(a+b)⁴ = 4C0 a⁴ b⁰+4C1a³b¹ +4C2 a²b² +4C3 a¹b³+4C4 a⁰b⁴
= a⁴ b⁰+4C1a³b¹+4C2 a²b² +4C3 a¹b³ +b⁴ (4C0=4C1=1)
(a+b)⁴= a⁴ +4a³b¹ +6a²b² +4a¹b³ +b⁴
Simillarly we can calculate
1) (a+b)⁵
= a⁵+5a⁴b¹ +10 a³b² +10a²b³ +5a¹b⁴+b⁵
2) (a+b)⁶
=a⁶+6a⁵b¹+15a⁴b²+20a³b³+15a²b⁴+6a¹b⁵+b⁶
3) (a+b)⁷
=a⁷+7a⁶b¹+21a⁵b²+35a⁴b³+35a³b⁴+21a²b⁵+7a¹b⁶+b⁷
This way we can calculate any power of (a+b) by using Binomial Expansion.
Note :
1) In the expansion of (a+b)ⁿ contains (n+1) terms .
For Example (a+b)² contain 3 terms , (a+b)³ contains 4 terms & so on .
Expansion contain one more term than the power of expansion.
2) First term is alway aⁿ & last term is bⁿ .
For Example in the expansion of (a+b)² first term is a²
and last term is b².
In Example of (a+b)³ , First term is a³ and last term is b³ so on .
3) In each term , the sum of power of each term
i.e a &b is alway n
For example in the expansion of (a+b)², power of 1st term is 2 ,
in second term sum of power of a & b is 2 and
Power of last term is also 2.
simillarly in (A+b)³& so on
4) In the successive term in the expansion we can
see that power of a decreases by 1 and Power of b increases by one.
Binomial Theorem can used for negative term also .
For example
(a-b)⁴= a⁴-4a³b¹+6 a²b² - 4a¹b³ +b⁴
(a-b)⁵= a⁵-5a⁴b¹+10 a³b²-10a²b³ +5a¹b⁴- b⁵
Here we can note that First term is +ve ,
second term is negative & so on. i.e.
Alternate terms are +ve & -ve .
Binomial can also used for fractional & negative powers also .
1/1+x = (1+x)⁻¹ =1-x+x²-x³+x⁴-x⁵+.......
1/1-x =(1-x)⁻¹ = 1+x+x²+x³+x⁴+x⁵+.......
1/(1+x)² = (1+x)⁻² =1-2x+3x²-4x³+........
1/(1-x)² =(1-x)⁻² =1+2x+3x²+4x³+......
1/√(1+x) =(1+x)⁻¹/² = 1+x/2-x²/8+x³/16+........
1/√(1-x) =(1-x)⁻¹/² = 1-x/2-x²/8-x³/16-.........
READ THE BLOG ON REPUBLIC DAY 26 JAN
Sunday, 1 November 2020
ANGLE & IT'S TYPES
ANGLE & IT'S TYPES
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