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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)ⁿis ⁿC_r a^n-rx^r

The (r+1)th term in the binomial expansion of (a-x)ⁿ is (-1)^{r n}C_ra^n-rx^r

Total number of terms in the binomial expansion:

Total number of terms in the binomial expansion of (a+x)ⁿor (a-x)ⁿis 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,

ⁿC_ma^n-mx^m= ⁿC_n/2a^n/2x^n/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 ⁿC_ma^n-mx^m

= ⁿC_{(n-1) / 2}a^{(n+1) / 2}x^{(n-1) / 2}

The (m+2)th term in the binomial expansion is ⁿC_m+1a ^{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 n}C_n/2a^n/2x^n/2when n is an even positive integer

(-1)^{(n-1)/2 n}C_{(n-1) / 2}a^{(n+1) / 2}x^{(n-1) / 2}or (-1)^{(n+1) / 2 n}C_{(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)ⁿwhere 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 n}C_{(n+1) / 2}a^{(n-1) / 2}x^{(n+1) / 2}has the same sign as 'a' & the middle term

(-1)^{(n-1)/2 n}C_{(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 n}C_{(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 n}C_{(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,

[ ⁿC_ra^n-rx^r ] / [ ⁿC_r-1a^n-r+1x^r-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<............<ⁿC_{n / 2}> ⁿC_{n / 2 + 1}> ...............>ⁿC_{n - 1}>ⁿC_n

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<............<ⁿC_{n-1 / 2}= ⁿC_{n+1 / 2}> ...............>ⁿC_{n - 1}>ⁿC_n

The above expression reminds us of a pyramid with flat top, much like a plateau.

Now, ⁿC₀= ⁿC_n

ⁿC₁ = ⁿC_{n - 1}

ⁿC₂ = ⁿC_{n - 2}

In general, ⁿC_r = ⁿC_{n - r}

So, we can conclude from the above facts, that,

1. The middle term in the expansion of (1+x)ⁿwhere 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)ⁿwhere n is an odd positive integer are the greatest coefficients of x.

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