Answer:
Binomial expansion also known as binomial theorem is a quick and easy way of expanding by multiplying binomial expression that has been raised to large power.
Step-by-step explanation:
A binomial is a polynomial with two terms. From the above definition given, if for instance an expression of (5x-4)²² was given, it wouldn't be that easy to multiply out by hand which is why the binomial theorem formula was created.
Another example to buttress this point is if we have (a³+2ab+b2)(a+b) expanding this will give a³+3a2b+b³ = a³+3a2b) +3ab²+b³)a+b = a⁴+ 4a³b+ 6a²b²+4ab³+b4..
This calculation could go on and on because it gets longer as we go but to make it easier and simpler, the best method to use here is the binomial theorem or expansion and the formula states thus:
n. n
(x+y)ⁿ = € {n}xⁿ-k*y^k
k=0 {k} = € {n} x^k*y^n-k
k=0 {k}
{n}
{k}. = binomial coefficient of each term
It should from the example given that we have index power(power raised)called exponent such that when an exponent is 0, we get 1 , i.e anything raised to power 0 =1
e.g (x+y)^0 = 1
And also when an exponent is 1, then the answer remain unchanged such that (x+y)¹= x+y. An exponent of 2 means to multiply given term by itself in two times and so on.
So to expantiate an expression like
a³+3a²b+3ab²+b³.. This expression has its index power(exponent) start with 3 such that .
The a goes down in the form.. 3,2,1,0 while the exponents of b go upwards in the form 0,1,2,3coeffiThe terms 0-n goes thus
k=0 k=1 k=2 k=3
a³ a² a 1
1 b. b² b³
which equals an-kbk.. This is confusing right. This is how it works
k=0 k=1 k=2 k= 3
an-kb^k. an-kb^k an-kb^k an-k^bk
a³-0b^0. a3-1b¹ a3-2b2 a3-3b3
a³ a²b. ab² b³
So when we put all this together we have: a³+a²b+ ab²+ b³. Actually this answer is correct but what we are missing here is the coefficient of n because we needs to include it.
So this becomes a³+3a²b +3ab²+b³ , the coefficient of n here is 3.
Now if we are to look at the result we got before from(a+b)^0 up to (a+b)³
And here that we have 3 as coefficient but where it is not shown we put 1 which is.
1
1a+ 1b
1a+2ab+ 1b²
1a³ +3a²b+3ab²+1b³... We solve this using pascal triangle.
1 This is solved by adding
1 1 each two numbers above
1 2 1 together to get the one
1 3 3 1 below.
1 4 6 4 1
This pascal pattern can be used for any exponents..1,2,3,5....60...120....n
Binomial theorem is basically the essence of this Pascal triangle pattern.
As stated earlier that the formula of binomial theorem is
n
€{n}a^n-kb^k
n-1 where €= summation
n= numbers given
k=coefficient of n
Now, let's to put into practical by solving this expression.
(x+y)ⁿ=€ⁿ {n}a^n-k* b^k
k=0 {k}
But before we solve this, its important we know the basis of Pascal triangle by using factorial to solve it with the formula n!/k!(n-k)! Such that
Factorial is n multiplied in descending order(i.e multiplying Numbers between 1-n)
K- coefficient of n
N- product of all the numbers between 1&n. For instance we are given this 4!/3!(4-3)! = 4!/3!*1!
= 4*3*2*1/3*2*1 *1 = 24/6 = 4
Now let's go back to the previous example of (x+y)ⁿ where n=4
(x+y)⁴= €⁴ {4} x^4-k*y^k
k=0 {k}
(x+y)⁴= €{4} (x^4-0)y^0 + {4} (x^4-1) *y¹
{0} {1}
+{4} (x^4-2)*y² + {4} (x^4-3)*y³
{2} {3}
+ {4} (x^4-4) * (y⁴)
{4}
= 1*x⁴y^0 + 4*x³y¹ + 6*x²y² +4*x¹y³ +1*x^0y⁴=
x⁴+4x³y +6x²y² + 4xy³ +y⁴(Final answer)
Please note that:
When solving binomial theorem expression, the easiest way to do this is by remembering the patterns below:
* first term's exponent start at n and go down
* second term's exponent start at 0 and goes up
* coefficient are solved with pascal triangle using n!/k!(n-k)!
on the line to the origin is given by
and 
attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.
, you find a critical point of 
.
