( a²+ b² +2ab ) - ( a²+ b²- 2ab )
Answers
Answer:
As to explain this we are Considering a binomial..
i.e…..(a+b)…
First.... Lets take it with 0 as exponent..
i.e…(a+b)^0
Which is equal to 1…
So, (a+b)^0=1…
Now, We r taking...
(a+b)^1= a+b
(a+b)^2=a^2 + 2ab + b^2….(this is your quarry... Wait i m going to solve it…but for now keep analyzing what i m doing)
(a+b)^3=a^3 + 3a^2b+ 3ab^2 + b^3…
It will be going on,
Now.. Here's the analyzing part.... U can analyze clearly that.. The first and the last coefficient of each digit is 1 but middle terms coefficient are continuesly changing.... We have to find a relation for those....
Now.. For that i m making a triangle.... Called... “Pascal Triangle" see this..
[(a+b)^0]........................1
[(a+b)^1]............1........................1
[(a+b)^2]..1..…………………x.....................…….1
………1……………….x………………………..x......................…1
[ABOVE WAS FOR (a+b)^3]
[ here dots doesn't have there usual meaning... I m drawing them for spacing... Consider only “1” and “x"]
Note: here, 1= coefficient of first and last term
X=coefficient of middle terms
Now.... To find the value of x.... We have to add the above terms over x...
For example….the value of x in (a+b)^2 will be 2..(1+1)
From this U can easily find out why coefficient of middle terms of (a+b)^3 is 3…[1+2=3 and 2+1=3]
So by this we can derive a formulae as...
(a+b)^n = [1]a^(n)b^(0) + [x]a^(n-1)b^(1) + [x]a^(n-2)b(2)…………+ [x]a^(1)b^(n-1) + [1]a^(0)b(n)
Where.... n=highest power , x= coeff. of middle terms
U can see that the power of “a" is gradualy decreasing until it becomes 0 while of “b" is increasing simantanously....
now... We are replacing the 1s(coeff. of first and last terms) and the x(coeff of middle terms) from the pascal triangle by “nC0” as its value is “1” and “nCn" respectively...
Therefore , a new formulae is in front of us.. i.e
(a+b)^n = nC0(a)^n(b)0 + nC1(a)^n-1(b)1 + ……………..+ nCn(a)^1(b)^n-1 + nCn(a)0(b)^n
I THINK... NOW ITS EASY TO FIND WHY
(a+b)^2 = a^2 + 2ab + b^2