# Binomial Theorem For Positive Integer Indices

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Let us now consider more formally the binomial theorem. We need to expand \({\left( {x + y} \right)^n}\) , where *x*, *y* are two arbitrary quantities, but *n* is a positive integer.

In particular, if we write

\[{\left( {x + y} \right)^n} = \left( {x + y} \right)\left( {x + y} \right)\left( {x + y} \right)...\left( {x + y} \right) \qquad \left( {n\,\,{\rm{times}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...{\rm{ }}\left( 1 \right)\]

we need to find out the coefficient of \({x^i}{y^j}\) . Note that \(i + j\) ** must ** always equal *n*, so that we can write a general term of the expansion (without the coefficient) as \({x^r}{y^{n - r}}\) so that \(0 \le r \le n\) .

Now, to find the coefficient of \({x^r}{y^{n - r}}\) , note that we need the quantity *x*, *r* times, while *y* is needed *n *– *r* times. Thus, in (1), \({x^r}{y^{n - r}}\) will be formed whenever *x* is ‘contributed’ by *r* of the binomial terms, while *y* is ‘contributed’ by the remaining *n* - *r* of the binomial terms. For example, in the expansion of \({\left( {x + y} \right)^5}\) , to form \({x^2}{y^3}\) , we need *x* from 2 terms and *y* from 3:

**Fig. 23**

How many ways are there to form \({x^2}{y^3}\) ? In other words, how many times will \({x^2}{y^3}\) be formed? The number of times \({x^2}{y^3}\) is formed is what is the coefficient of \({x^2}{y^3}\) . That number, which would be immediately obvious to the alert reader, is simply \(^5{C_2}\) . Why? Because this is the number of ways in which we can select any 2 binomial terms from 5. These 2 terms will contribute *x*. The remaining will automatically contribute *y*.

In the general case of \({\left( {x + y} \right)^n}\) , we see that the coefficient of \({x^r}{y^{n - r}}\) would be \(^n{C_r}\) . (which is infact the same as \(^n{C_{n - r}}\) ). Thus, the general binomial expansion is

\[ \Rightarrow \qquad \fbox{${{{\left( {x + y} \right)}^n} = \sum\limits_{r = 0}^n {{\,^n}{C_r}} \,{x^{n - r}}{y^r}}$}\]

The coefficients \(^n{C_i}\) are called the binomial coefficients, for a reason that should now be obvious.

Note that the \({\left( {i + 1} \right)^{{\rm{th}}}}\) coefficient in this expansion is \(^n{C_i}\) , which now explains the relation

\[{T_{n,\,i + 1}} = {T_{n - 1,i}} + {T_{n - 1,\,i + 1}}\]

we observed in the Pascal triangle; this relation simply corresponds to

\[^n{C_i} = {\,^{n - 1}}{C_{i - 1}} + {\,^{n - 1}}{C_i}\]

Also, the binomial coefficients of terms equidistant from the beginning and the end are equal, because we have \(^n{C_r} = {\,^n}{C_{n - r}}\) . The general term of expansion, \(^n{C_r}{x^{n - r}}{y^r}\) , is the (*r* + 1) th term from the beginning of the expansion and is conventionally denoted by \({T_{r + 1}}\) , i.e.

\[{T_{r + 1}} = {\,^n}{C_r}\,{x^{n - r}}{y^r}\]

Since we have (*n* + 1) terms in the general expansion, we see that if *n* is even, there will be an odd number of terms, and thus there will be only one middle term, which would be \(^n{C_{n/2}}\,{x^{n/2}}{y^{n/2}}\) . For example,

\[{\left( {x + y} \right)^4} = {x^4} + 4{x^3}y + \mathop {6{x^2}{y^2}}\limits_{\scriptstyle\,\,\,\,\,{\rm{only}}\,{\rm{one}} \atop\scriptstyle{\rm{middle}}\,{\rm{term}}} + 4x{y^3} + {y^4}\]

On the other hand, if *n* is odd, then there will be an even number of terms in the expansion, and thus there will be two middle terms, namely \(\begin{align}^n{C_{\frac{{n - 1}}{2}}}\,\,{x^{\frac{{n + 1}}{2}}}\,{y^{\frac{{n - 1}}{2}}}\,\,{\rm{and}}\,{\,^n}{C_{\frac{{n - 1}}{2}}}\,\,{x^{\frac{{n - 1}}{2}}}\,{y^{\frac{{n + 1}}{2}}}\end{align}\)

For example;

\[{\left( {x + y} \right)^5} = {x^5} + 5{x^4}y + \mathop {10{x^3}{y^2} + 10{x^2}{y^3}}\limits_{{\rm{Two}}\,\,{\rm{middle}}\,\,\,{\rm{terms}}} + 5x{y^4} + {y^5}\]

**Example – 1**

Find the middle term(s) in the expansion of \({\left( {x - \frac{1}{x}} \right)^9}\)

**Solution: ** Since * n * = 9, there will be 10 terms in the expansion, which means that there will be 2 middle terms in the expansion, the 5 th and the 6 th :

\[\begin{array}{l}{T_5} = {\,^9}{C_4}{\left( x \right)^{9 - 4}}\left( {\frac{{ - 1}}{x}} \right)_{}^4 = 126\,x\\{T_6} = {\,^9}{C_5}{\left( x \right)^{9 - 5}}{\left( {\frac{{ - 1}}{x}} \right)_5} = \frac{{ - 126}}{x}\end{array}\]

**Example – 2**

Is there any term in the expansion of \({\left( {\frac{{2{x^2}}}{5} - \frac{1}{{\sqrt x }}} \right)^{10}}\) that will be independent of *x*?

**Solution: ** The general term in the expansion is

\[\begin{array}{l}{T_{r + 1}} = {\,^{10}}{C_r}{\left( {\frac{{2{x^2}}}{5}} \right)^{10 - r}}{\left( {\frac{{ - 1}}{{\sqrt x }}} \right)^r}\\\,\,\,\,\,\,\,\,\, = {\,^{10}}{C_r}{\left( {\frac{2}{5}} \right)^{10 - r}}{\left( { - 1} \right)^r}\,\,{x^{20 - 2r - \frac{r}{2}}}\\\,\,\,\,\,\,\,\,\, = {\,^{10}}{C_r}{\left( {\frac{2}{5}} \right)^{10 - r}}{\left( { - 1} \right)^r}\,\,{x^{20\, - \frac{{5r}}{2}}}\end{array}\]

Thus, for the term that is independent of *x*, we have

\[\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,20 - \frac{{5r}}{2} = 0\\ \Rightarrow \qquad r = 8\end{array}\]

Thus, the term free of *x* is the 9 ^{th} term given by

\[{T_9} = {\,^{10}}{C_8}{\left( {\frac{2}{5}} \right)^2}{\left( { - 1} \right)^8} = \frac{{36}}{5}\]

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school