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The expression that consists of two terms is known as a binomial expression. Example :
$$\;a\;+\;b,\;x+\frac{1}{y}\;,\;x^2+y\;$$ . When the power of such expressions is a very small positive integer such as 2, 3, 4 it is not difficult to expand them. But if the power is large numbers, let us say ‘n’ then we need a formula a formula for the expansion of the binomial expression. The expansion of the [removed]a+x)^{n}, where the value of n is a positive integer is called the Binomial Theorem. The binomial theorem was first introduced by Sir Issac Newton.
According to the binomial theorem, for any positive integer n,
$$(a+x)^n\;=\;^nC_0a^nx^0\;+\;^nC_2a^{n-1}x^1\;+\;^nC_3a^{n-2}x^2\;+\;.\;.\;.\;.\;.\;.\;+\;^nC_ra^{n-r}x^r\;+\;.\;.\;.\;.\;.\;+\;^nC_na^{n-n}x^n$$
Proof: We have, $$\;(a+x)\;=\;(a+X)(a+x)(a+x)\;.\;.\;.\;.\;.\;.\;.\;to\;n\;factors$$
In the process of multiplication of n factors, let us start by taking n = 2, 3, and 4. Then the expansions are :$$\;(a+x)^1=a+x$$ $$\;(a+x)^2=a^2+2ab+b^2$$ $$\;(a+x)^3=a^3+3a^2b+3ab^2+b^3$$ $$\;(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
Form the above equation we can see that,
$$\;(a+x)^2=a^2+2ab+b^2\;=^2C_0a^2b^0+^2C_1ab+^2C_2a^0b^2$$
$$\;(a+x)^3=a^3+3a^2b+3ab^2+b^3\;=^3C_0a^3b^0+^3C_1a^2b^1+^3C_2a^1b^2+^3C_3b^3$$
So, the theorem holds good when n= 2, 3.
Let us assume that the theorem is true for all positive integer values of n. Let us say m, is the new value of n, then
$$(a+b)^m\;=\;^mC_0a^mb^0\;+\;^mC_1a^{m-1}b^1\;+\;^mC_2a^{m-2}b^2\;+\;.\;.\;.\;.\;.\;.\;+\;^mC_ra^{m-r}b^r\;+\;.\;.\;.\;.\;.\;+\;^mC_ma^{m-m}b^m$$
Multiplying both sides by (a + b), we get
$$(a+b)^m+1\;=\;(a+b)\;(\;^mC_0a^mb^0\;+\;^mC_1a^{m-1}b^1\;+\;^mC_1a^{m-2}b^2\;+\;.\;.\;.\;.\;.\;.\;+\;^mC_ra^{m-r}b^r\;+\;.\;.\;.\;.\;.\;+\;^mC_ma^{m-m}b^m\;)$$
$$\;=\;^mC_0a^{m+1}b^0\;+\;^mC_1a^{m}b^1\;+\;^mC_2a^{m-1}b^2\;+;.\;.\;.\;.\;.\;+\;^mC_ma^{1}b^m\;+\;.\;.\;.\;.\;.\;.\;^mC_0a^mb^1\;+\;^mC_1a^{m-1}b^2\;+\;^mC_1a^{m-2}b^3\;+\;.\;.\;.\;.\;+\;^mC_mb^{m+1}$$
$$\;=\;a^{m+1}\;+\;(^mC_1\;+\;1)\;a^mx\;+(^mC_2\;+\;^mC_1\;)\;a^{m-1}x^2\;+\;(^mC_3\;+\;^mC_2\;)\;a^{m-2}x^3\;+\;.\;.\;.\;.\;.\;+\;1\;x^{m+1}$$
$$\;=\;^{m+1}C_0a^{m+1}\;+\;^{m+1}C_1a^mx\;+\;^{m+1}C_2a^{m-1}x^2\;+\;.\;.\;.\;.\;.\;.\;+\;^{m+1}C_{m+1}x^{m+1}$$
Therefore when the theorem is true for n = m. The theorem is also true for n = m + 1.
We know that the theorem is true for n = 3, therefore it must be true n = 4
Some of the important properties on binomial expansion are as follows :
The first term T_{1} = T_{0 + 1} = ^{n}C_{0} a^{n-0} b^{0}
The second term T_{2} = T_{1 + 1} = ^{n}C_{1} a^{n-1} b^{1}
The third term T_{3} = T_{2 + 1} = ^{n}C_{2} a^{n-2} b^{2} . . . . . . . . . . .
Similarly, the (r + 1)^{th} term T_{r + 1} = ^{n}C_{r} a^{n-r} b^{r} which is called the general term.
Now, let us find the middle term or terms of the expansion (a + b)^{n}. We have to consider the cases when n is an even number and when n is an odd number.
When n is even, we write n = 2m ,where m = 1, 2, 3, . . . . The number of terms after expansion is 2m + 1, which is odd. So, it has only one middle term, namely (m + 1)th term. So,
$$\;t_{m+1}\;=\;^{2m}C_ma^mb^m\;=\;\frac{2m!}{(m!)^2}\;a^m\;b^m$$
$$\therefore\;middle\;term\;t_{m+1}\;=\;t_{\frac{1}{2}n+1}$$
$$\;=\;^nC_{\frac{1}{2}n}a^{n-\frac{n}{2}}x^{\frac{n}{2}}$$
$$\;=\;\frac{n!}{(\frac{1}{2}n!)^2}a^{n-\frac{n}{2}}x^{\frac{n}{2}}$$
When n is odd, we write n = 2m – 1, where m = 1, 2, 3 …., The number of terms after expansion is 2m, which is even. So, there will be two middle terms, namely m^{th} and (m + 1)th term. So,
$$\;t_m\;=\;^{2m-1}C_{m-1}.a^m.x^{m-1}$$
$$\;=\;\frac{(2m-1)!}{m!(m-1)!}\;.a^m.x^{m-1}$$
and,
$$\;t_{m+1}\;=\;^{2m-1}C_m.a^{m-1}.x^m$$
$$\;=\;\frac{(2m-1)!}{m!(m-1)!}\;.a^{m-1}.x^{m}$$
$$\therefore\;the\;middle\;terms\;are$$
$$\;t_m\;=\;t_{\frac{n+1}{2}}\;=\;t_{\frac{n-1}{2}+1}$$
and,
$$t_{m+1}\;=\;t_{\frac{n+1}{2}+1}$$
Taken reference from
( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )
According to the binomial theorem, for any positive integer n,
$$(a+x)^n\;=\;^nC_0a^nx^0\;+\;^nC_2a^{n-1}x^1\;+\;^nC_3a^{n-2}x^2\;+\;.\;.\;.\;.\;.\;.\;+\;^nC_ra^{n-r}x^r\;+\;.\;.\;.\;.\;.\;+\;^nC_na^{n-n}x^n$$
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