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Binomial Theorem
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Rule
Binomial Theorem
Consider the power of a binomial (x+y)^n, where n is a natural number. The binomial expansion can be written as a sum of terms.
Each of the terms _nC_j are the combinations of n objects taken j at a time. The values of _nC_j are the same as those in the nth row of Pascal's triangle. Therefore, the binomial expansion can also be written in terms of the Pascal's Triangle's values of the nth row.
Proof
To prove this theorem, a method called Mathematical Induction will be used. First the small cases need to be examined. Consider that n=1. (x+y)^1 = x + y
Since _1C_0 and _1C_1 are both equal to 1, the theorem holds for this case. Now consider n=2.
(x+y)^2 = x^2 + 2xy + y^2
On this case, _2C_0 and _2C_2 are equal 1 and _2C_1 is equal to 2, making this case also true. Following this, it can be assumed that the theorem holds for a natural number k.
To prove the theorem, it needs to be shown that the theorem holds for the value of k+1. This can be done by multiplying the case above by (x+y). (x+y)^(k+1)=(x+y)(x+y)^k To expand the binomial, multiply (x+y) by the binomial expansion of (x+y)^k. This has to be done carefully, as distributing x and y modify the terms on the binomial expansion.
Examining the resulting terms closely, it can be noted that there are terms that can be factored together. For example, consider the following. _kC_1x^ky^1 + _kC_0 x^k y^1 &= ( _kC_0 + _kC_1)x^ky^1 _kC_2x^(k-1)y^2 + _(k-1)C_1 x^(k-1) y^2 &= ( _kC_1 + _kC_2)x^(k-1)y^2 Using these examples, it is possible to find a way to write these terms. Let r be a natural number from 1 to k. Using this variable, factored terms can be written as follows. (_kC_(r-1) + _kC_r)x^(k-r+1)y^r Now the sum between parenthesis will be examined to see if it can be simplified. The expressions for the combinations will be used to do so.
_kC_(r-1) + _kC_r
▼
Simplify
SubstituteExpressions
Substitute expressions
k!/(k-(r-1))!(r-1)! + k!/(k-(r))!(r)!
FactorOut
Factor out k!
k!(1/(k-(r-1))!(r-1)! + 1/(k-(r))!(r)! )
DistrNegSignSwap
-(b-a)=a-b
k!(1/(k+1-r)!(r-1)! + 1/(k-r)!(r)! )
ExpandFrac
a/b=a * r/b * r
k!(r/(k+1-r)!(r-1)!r + 1/(k-r)!(r)! )
(r-1)!r = r!
k!(r/(k+1-r)!(r)! + 1/(k-r)!(r)! )
ExpandFrac
a/b=a * (k+1-r)/b * (k+1-r)
k!(r/(k+1-r)!(r)! + k+1-r/(k-r)!(r)!(k+1-r) )
(k-r)!(k+1-r)=(k+1-r)!
k!(r/(k+1-r)!(r)! + k+1-r/(k+1-r)!(r)! )
AddFrac
Add fractions
k!( k+1/(k+1-r)!(r)! )
Multiply
Multiply
k!(k+1)/(k+1-r)!(r)!
k!(k+1) = (k+1)!
(k+1)!/(k+1-r)!(r)!
SubstituteExpressions
Substitute expressions
_(k+1)C_r
Considering this result, the repeating terms can be rewritten to simplify the binomial expansion.
Looking at the binomial expansion, it can be noted that the repeated terms are written like the theorem. The remaining terms x^(k+1) and y^(k+1) can be rewritten considering the values of _kC_0 x^(k+1) and ·_kC_k x^0y^(k+1).
| Combination | Formula | Simplify |
|---|---|---|
| _kC_0 | k!/(k-0)!0! | 1 |
| _kC_k | k!/(k-k)!k! | 1 |
| _(k+1)C_0 | (k+1)!/(k+1-0)!0! | 1 |
| _(k+1)C_(k+1) | (k+1)!/(k+1-(k+1))!(k+1)! | 1 |
Since of the values on the table are equal to 1, these can be interchanged to rewrite the expression one last time.
Therefore, the result holds for in the case that n=k+1, which finish the proof.
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