An algebraic expression containing two terms is called binomial expression.
Example
$( x+a)$
$(\frac {1}{2} +x)$
$(\frac {2}{x} - \frac{1}{x^3})$
The general form of the binomial expression is $(x+a)$ and expansion $(x+a)^n , n \in N$ is called Binomial expression
It was developed by Sir Issac newton
The general expression for the Binomial Theorem is
$(x+a)^n = ^{n}C_{0} x^n a^0 + ^{n}C_{1} x^{n-1} a^1 + ^{n}C_{2} x^{n-2} a^2 + .....+ ^{n}C_{r} x^{n-r} a^r ....+^{n}C_{n} x^{0} a^n $
$(x+a)^{n} = \sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^{k} $ Proof:
We can prove this theorem with the help of mathematical induction
Let us assume P(n) be the statement is
$(x+a)^n = ^{n}C_{0} x^n a^0 + ^{n}C_{1} x^{n-1} a^1 + ^{n}C_{2} x^{n-2} a^2 + .....+ ^{n}C_{r} x^{n-r} a^r ....+ ^{n}C_{n} x^{0} a^n $ Step 1
Now the value of P(1)
$(x+a)^1 = ^{1}C_{0} x^1 a^0 + ^{1}C_{1} x^{1-1} a^1$
$=(x+a)$
So P(1) is true Step 2
Now the value of P(m)
$(x+a)^m = ^{m}C_{0} x^m a^0 + ^{m}C_{1} x^{m-1} a^1 + ^{m}C_{2} x^{m-2} a^2 + ...+ ^{m}C_{m} x^{0} a^m $
Now we have to prove
$(x+a)^{m+1} = ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 +...+ ^{m+1}C_{m+1} x^{0} a^{m+1} $
Now
$(x+a)^{m+1} =(x+a)(x+a)^m $
$= (x+a)( ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 + ...+ ^{m+1}C_{m+1} x^{0} a^{m+1} $
$= ^m C_0 x^{m+1}a{0} + ( ^mC_1 + ^mC_0) x^ma^1 + (^mC_2 + ^mC_1) x^{m-1}a^2 +.... $
$+(^m C_{m-1} + ^mC_{m}) x^1a^m + ^mC_m x^0a^{m+1}$
As $ ^mC_{r-1} + ^mC_r = ^{m+1}C_r$
So,
$= ^{m+1}C_{0} x^{m+1} a^0 + ^{m+1}C_{1} x^{m} a^1 + ^{m+1}C_{2} x^{m-1} a^2 + ...+ ^{m+1}C_{m+1} x^{0} a^{m+1} $
So by principle of Mathematical induction, P(n) is true for $n \in N$
Important conclusion from Binomial Theorem
1
$(x+a)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^{k} $
We can easily see that $(x+a)^n$ has $(n+1)$ terms as k can have values from 0 to n
2
The sum of indices of x and a in each is equal to n
$x^{n-k}a^{k}$
3
The coefficient nCr is each term is called binomial coefficient
4
$(x-a)^n$ can be treated as $[x+(-a)]^n$
So
$(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k $
So the terms in the expansion are alternatively positive and negative. The last term is positive or negative depending on the values of n
5.
$(1+x)^n$ can be treated as $(x+a)^1$ where x=1 and a=x
So
$(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$
This is the expansion is ascending order of power of x
6.
$(1+x)^n$ can be treated as $(x+a)^n$ where x=x and a=1
So
$(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}$
This is the expansion is descending order of power of x
7.
$(1-x)^n$ ;can be treated as $(x+a)^n$ where x=1 and a=-x
So
$(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$
This is the expansion is ascending order of power of x
$(x+a)^n =\sum_{k=0}^{n} {}^{n} C_k x^{n-k}a^k $
The First term would be = $ ^{n} C_0 x^n a^0$
The Second term would be =$ ^{n} C_1 x^{n-1} a^1$
The Third term would be = $ ^{n} C_2 x^{n-2} a^2$
The Fourth term would be = $ ^{n} C_3 x^{n-3} a^3$
Like wise (k+1) term would be
$T_{k+1}= ^n C_k x^{n-k}a^k $
This is called the general term also as every term can be find using this term
$T_1= T_{0+1}=^n C_0 x^n a^0 $
$T_2= T_{1+1}=^n C_1 x^{n-1} a^1 $
Similarly for
$(x -a)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{n-k}a^k $
$T_{k+1}= ^n C_k (-1)^k x^{n-k}a^k $
Again similarly for
$(1 +x)^n =\sum_{k=0}^{n} {}^{n} C_k x^{k}$
$T_{k+1}= ^n C_k x^{k}$
Again similarly for
$(1 -x)^n =\sum_{k=0}^{n} {}^{n} C_k (-1)^k x^{k}$
$T_{k+1}= ^n C_k (-1)^k x^{k}$ To summarize it
If n is even then the middle term would $[(\frac {n}{2}+1 ]$ th term
If n is odd,then $\frac{n+1}{2} $ and $ \frac {n+3}{3}$ are the middle term
Solved Examples
Question-1
If the coefficient of $(2k + 4)$ and $(k - 2)$ terms in the expansion of $(1+x)^{24}$ are equal then find the value of k Solution:
The general term of $(1 + x)^n$ is
$T_{k+1}= ^n C_k x^{k}$
Hence coefficient of (2k + 4)th term will be
$T_{2k+4} = T_{2k+3+1} = ^{24} C_{2k+3}$
and coefficient or (k - 2)th term will be
$T_{k-2} = T_{k-3+1} = ^{24} C_{k-3}$
As per question both the terms are equal 24C2k+3 = 24Ck-3.
or (2k + 3) + (k-3) = 24
k = 8 Practice Question
