BINOMIAL THEOREM
Consider the following expression :
( a + x )n= ( a + x )( a + x ).............. n times ( where n is a positive integer )
First take all a's from the n bracketed terms and no x's from any other terms. This can be done only in 1 way or nC0 way.
Take one x from the ' n ' bracketed terms and multiply it by the remaining (n-1) a's from the other (n-1) bracketed terms. Now one x can be selected from the n bracketed terms in nC1 ways.
Similarly, take two x from the ' n ' bracketed terms and multiply it by the remaining (n-2) a's from the other (n-2) bracketed terms. Now two x can be selected from the n bracketed terms in nC2 ways. .............
Finally, Take all x from the ' n ' bracketed terms and no a's from any of the bracketed terms. Now n 'x' can be selected from the n bracketed terms in nCn ways.
So, coefficient of x^0 is nC0
coefficient of x^1 is nC1
in general, coefficient of x^r is nCr
and coefficient of x^n is nCn
Therefore, (a+x)(a+x)....... n times
= nC0 a^n + nC1 a^(n-1) x + nC2 a^(n-2) x^2 + .....+nCr a^(n-r) x^r + ...nCn x^n
i.e (a+x)^n
= nC0 a^n + nC1 a^(n-1) x + nC2 a^(n-2) x^2 + .....+nCr a^(n-r) x^r + ...nCn x^n
= a^n + nC1 a^(n-1) x + nC2 a^(n-2) x^2 + .....+nCr a^(n-r) x^r + ...... x^n
Try proving the above formula by the first principle of Mathematical induction.
Now let us obtain some corollaries with this formula, given above.
COROLLARY 1:
Put '-x' instead of 'x'
(a-x)^n
= a^n - nC1 a^(n-1)x + nC2 a^(n-2)x^2 - ........ (-1)^r nCr a^(n-r) x^r .....
(-1)^n x^n
COROLLARY 2:
Put a = x = 1 in the original formula,
nC0 + nC1 + nC2 + ..................... + nCn = ( 1 + 1 )^n = 2^n
COROLLARY 3:
Put a = x = 1 in corollary 1,
1 - nC1 + nC2 - .....(-1)^r nCr ...(-1)^n x^n = (1-1)^n = 0
i.e nC0 + nC2 + nC4 +.........nCk = nC1 + nC3 + nC5 + ....... nCm
[ where,
k = n when n is even positive integer
= n-1 when n is odd positive integer
&
m = n when n is odd positive integer
= n-1 when n is even positive integer ]
i.e sum of coefficients of even powers of x in the binomial expansion of (1+x)^n
= sum of coefficients of odd powers of x in the binomial expansion of (1+x)^n
COROLLARY 4:
From corrolary 3, let,
nC0 + nC2 + nC4 +.........nCk = nC1 + nC3 + nC5 + ....... nCm = S
[ where,
k = n when n is even positive integer
= n-1 when n is odd positive integer
&
m = n when n is odd positive integer
= n-1 when n is even positive integer ]
Using corrolary 2,
2^n = nC0+nC1+nC2+........+nCr+.......nCn = 2S
i.e S = 2^(n-1)
Hence,
sum of coefficients of even powers of x in the binomial expansion of (1+x)^n
= sum of coefficients of odd powers of x in the binomial expansion of (1+x)^n
= 2^(n-1)
COROLLARY 5:
(1+x)^n = nC0 + nC1x + nC2x^2 + .........+ nCrx^r+..............nCnx^n
(x+1)^n = nC0x^n + nC1x^(n-1) + ...........nCr x^(n-r)+.........nCn
Multiplying the above two expressions and comparing the coefficient of x^n,
2nCn = (nC0)^2 + (nC1)^2 + (nC2)^2 + .............+(nCr)^2+.........+(nCn)^2
General Term in the binomial expansion:
The (r+1)th term in the binomial expansion of (a+x)n is nCr an-r xr
The (r+1)th term in the binomial expansion of (a-x)n is (-1)r nCr an-r xr
Total number of terms in the binomial expansion:
Total number of terms in the binomial expansion of (a+x)n or (a-x)n is n+1
THE MIDDLE TERMS IN THE BINOMIAL EXPANSION:
Case: 1
When n = 2m, m being a positive integer.
In this case, total number of terms in the binomial expansion is 2m + 1 i.e there are odd number of terms, hence there is one and only one middle term.
The (m+1)th term in the expansion is the middle term which is given by,
nCm an-m xm = nCn/2 an/2 xn/2
Case: 2
When n = 2m + 1, m being a positive integer.
In this case, total number of terms in the binomial expansion is 2m + 1 + 1 =
2m + 2 i.e there are even number of terms, hence there are two middle terms in the binomial expansion. The (m+1)th and (m+2)th terms are the middle terms in the binomial expansion.
The (m+1)th term in the binomial expansion is nCm an-m xm
= nC(n-1) / 2 a(n+1) / 2 x(n-1) / 2
The (m+2)th term in the binomial expansion is nCm+1 a n ? m -1 x m-1
= nC(n+1) / 2 a(n-1) / 2 x(n+1) / 2
Note:
In the expansion of (a-x)n the middle term(s) is(are)
(-1)n / 2 nCn/2 an/2 xn/2 when n is an even positive integer
(-1)(n-1)/2 nC(n-1) / 2 a(n+1) / 2 x(n-1) / 2 or (-1)(n+1) / 2 nC(n+1) / 2 a(n-1) / 2 x(n+1) / 2
when n is an odd positive integer.
Remark:
In the binomial expansion of (a-x)n where n is an even positive integer, the middle term is positive when n is a multiple of 4.
In the binomial expansion of (a-x)n where n is an odd positive integer, the two middle terms can never be of the same sign i.e one is positive and the other is negative.
Smashing Tips:
In the binomial expansion of (a-x)n where n = 2k, & k is an odd positive integer i.e n is an even integer and is a multiple of 2 and not of 4, then the middle term of the expansion is positive when a & x are of the opposite sign (either a or x is positive, the other is negative), and it is negative when a & x are of the same sign (both a & x are positive or negative).
Rocking Tips:
In the binomial expansion of (a-x)n where n is an odd positive integer of the form 4k+3; k being any positive integer, then the middle term of the expansion,
(-1)(n+1) / 2 nC(n+1) / 2 a(n-1) / 2 x(n+1) / 2 has the same sign as 'a' & the middle term
(-1)(n-1)/2 nC(n-1) / 2 a(n+1) / 2 x(n-1) / 2 has the opposite sign of that of 'x'.
In the binomial expansion of (a-x)n where n is an odd positive integer of the form 4k+1; k being any positive integer, then the middle term of the expansion,
(-1)(n+1) / 2 nC(n+1) / 2 a(n-1) / 2 x(n+1) / 2 has the opposite sign of that of 'x' & the middle term
(-1)(n-1)/2 nC(n-1) / 2 a(n+1) / 2 x(n-1) / 2 has the same sign as that of 'a'.
The Greatest Terms in the Binomial Expansion:
In (a+x)n
(r+1)th term / rth term is > or < or = 1, according as,
[ nCr an-r xr ] / [ nCr-1 an-r+1 xr-1 ] > or < or = 1, according as,
(n-r+1)/r . (x/a) > or < or = 1, according as,
(n-r+1)x > or < or = ar
The case of Pyramidal Increase & Decrease with perfect symmetry:
Consider the binomial expansion of (1+1)n
(r+1)th term is > or < or = rth term, according as,
n - r + 1 > or < or = r, according as,
(n+1)/2 > or < or = r.
So, when n is an even positive integer, the binomial coefficients can be arranged according to the following,
nC0<nC1<nC2<............<nCn / 2 > nCn / 2 + 1 > ...............>nCn - 1 >nCn
The above expression reminds us of a pyramid with peaked top.
Again, when n is an odd positive integer, the binomial coefficients can be arranged according to the following,
nC0<nC1<nC2<............<nCn-1 / 2 = nCn+1 / 2 > ...............>nCn - 1 >nCn
The above expression reminds us of a pyramid with flat top, much like a plateau.
Now, nC0 = nCn
nC1 = nCn - 1
nC2 = nCn - 2
In general, nCr = nCn - r
So, we can conclude from the above facts, that,
1. The middle term in the expansion of (1+x)n where n is an even positive integer is the greatest coefficient of x.
2. All the coefficients of x equidistant from the middle term are equal.
3. Both the middle terms in the expansion of (1+x)n where n is an odd positive integer are the greatest coefficients of x.
