Saturday, 9 August 2014

CHAPTER 2 - Binomial Theorem for Positive Integral Index

Let us now consider more formally the binomial theorem. We need to expand {\left( {x + y} \right)^n}, where xy 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)\ldots \left( {x + y} \right) \ \left( {n\,\,{\rm{times}}} \right)\ldots(1)
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 xrtimes, 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:
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
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}}\hfill\atop  \scriptstyle{\rm{middle}}\,{\rm{term}}\hfill}  + 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 ^n{C_{\dfrac{{n - 1}}{2}}}\,\,{x^{\dfrac{{n + 1}}{2}}}\,{y^{\dfrac{{n - 1}}{2}}} and ^n{C_{\dfrac{{n - 1}}{2}}}\,\,{x^{\dfrac{{n - 1}}{2}}}\,{y^{\dfrac{{n + 1}}{2}}}
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}
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