To determine the greatest coefficient in the binomial expansion, (1+x)ⁿ, when n is a positive integer. Coefficient of 

(Tr+1/Tr) = Cr/Cr-1 = (n-r+1)/r = ((n+1)/r) - 1.

Now the (r+1)^th binomial coefficient will be greater than the rth binomial coefficient when, T_r+1 > T_r

=>     ((n+1)/r)-1 > 1 => (n+1)/2 >r.                                   ....... (1)

But r must be an integer, and therefore when n is even, the greatest binomial coefficient is given by the greatest value of r, consistent with (1) i.e., r = n/2 and hence the greatest binomial coefficient is ⁿC_n/2.

Similarly in n be odd, the greatest binomial coefficient is given when,

r = (n-1)/2 or (n+1)/2 and the coefficient itself will be ⁿC_(n+1)/2 or ⁿC_(n-1)/2, both being are equal

Note: The greatest binomial coefficient is the binomial coefficient of the middle term.

 Illustration:

        Show that the greatest the coefficient in the expansion of (x + 1/x)²ⁿ is (1.3.5...(2n-1).2ⁿ)/n!  .

Solution:

        Since middle term has the greatest coefficient.

        So, greatest coefficient = coefficient of middle term

        = ²ⁿC_n = (1.2.3...2ⁿ)/n!n! = (1.3.5...(2n-1).2ⁿ)/n!.

Numerically greatest term

To determine the numerically greatest term in the expansion of (a + x)ⁿ, where n is a positive integer.

Consider

Thus

Note : {((n+1)/r) - 1} must be positive since n > r. Thus T_r+1 will be the greatest term if, r has the greatest value as per the equation (1).

Illustration:

        Find the greatest term in the expansion of (3-2x)⁹ when x = 1.

Solution:

         T_r+1/T_{r = ((9-r+1)/r) . (2x/3) >1}

        i.e. 20 > 5r

        If r = 4, then T_r+1 = T_r and these are the greatest terms. Thus 4^th and 5^th terms are numerically equal and greater than any other term and their value is equal 3⁹ × ⁹C₃ × (2/3)³ = 489888.

Illustration:

        Find the greatest term in the expansion of (2 + 3x)⁹ if x = 3/2.

Solution:

        Here  T_r+1/T_r = ((n-r+1)/r)(3x/2) = ((10-r)/r)(3x/2), (where x = 3/2)

        = ((10-r)/r)(3x/2)(3/2) = ((10-r)/r).9/4 = (90-9r)/4r

        Therefore T_r+1 > T_r, if 90 - 9r > 4r.

        => 90 > 13r => r < 90/13 and r being an integer, r = 6.

        Hence T_r+1 = T₇ = T₆₊₁ = ⁹C₆ (2)³ (3x)⁶ = 3¹³.7/2.

Illustration:

Given that the 4^th term in the expansion of (2 + (3/8)x)¹⁰ has the maximum numerical value, find the range of values of x for which this will be true.

Solution:

        Given 4^th term in (2 + (3/8)x)¹⁰ = 2¹⁰ (1 + (3/16)x)¹⁰, is numerically greatest

        =>|T4/T3| > 1 and |T5/T4| < 1

        =>

        =>|x| > 2 and |x| < 64/21

        => x ε [-(64/21),-2]U[2,(64/21)].

Illustration:

        Find the greatest term in the expansion of √3(1 + (1/√3))²⁰.

Solution:

        Let r^th term be the greatest

        Since T_r/T_r+1 = (r/(21-r)).√3.

Now T_r/T_r+1 > 1 => r > 21/(√3+1)                                             ...... (1)

Now again T_r/T_r+1 = (r/(21-r)) √3 < 1 => r <  (22+√3)/(√3+1)               ...... (2)

from (1) and (2)

22/(√3+1) < r < (22+√3)/(√3+1)

=> r = 8 is the greatest term and its value is √3 . ²⁰C₇­ (1/√3)⁷ = ²⁰C₇  (1/27).

Particular Cases

We have (a + x)ⁿ = aⁿ + ⁿC₁ a^n-1 x + ⁿC₂ a^n-2 x^r +...+ xⁿ.            ...... (1)

(i)     Putting x = -x in (1), we get

        (a-x)ⁿ = aⁿ - ⁿC₁ a^n-1x + ⁿC₂ a^n-2 x² - ⁿC₂ a^n-3 x³ +...+ (-1)^{r n}C_r a^n-r x^r +...+ (-1)ⁿ xⁿ.

(ii)    Putting a = 1 in (1), we get,

        (1+x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² +...+ ⁿC_r x^r +...+ ⁿC_nxⁿ.                    ... (A)

(iii)    Putting a = 1, x = -x in (1), we get

        (1-x)ⁿ=ⁿC₀-ⁿC₁ x + ⁿC₂ x²-ⁿC₃x³ +... (-1)^{r n}C_r x^r +... (-1)ⁿ _nC_nxⁿ     ... (B)

Tips to Remember

(a)    T_r/T_{r+1 = ((n-r+1)/r ).(x/a)} for the binomial expansion of (a + x)ⁿ.

(b)    ⁽ⁿ⁺¹⁾C_r = ⁿC_r + ⁿC_r-1.

(c)    r ⁿC_r = n ^n-1C_r-1

(d)  

(e)    When n is even,

        (x + a)ⁿ + (x - a)ⁿ = 2(xⁿ + ⁿC₂ x^n-2 a² + ⁿC₄ x^n-4 a⁴ +...+ ⁿC_n a_n).

        When n is odd,

        (x + a)ⁿ + (x - a)ⁿ = 2(xⁿ + ⁿC₂ x^n-2 a² +...+ ⁿC_n-1 x a^n-1).

        When n is even

        (x + a)ⁿ - (x - a)ⁿ = 2(ⁿC₁ x^n-1 a + ⁿC₃ x^n-3 a³ +...+ ⁿC_n-1 x a^n-1).

        When n is odd

        (x + a)ⁿ - (x - a)ⁿ = 2(ⁿC₁ x^n-1 a + ⁿC₃ x^n-3 a³ +...+ ⁿC_n aⁿ).

Properties of Binomial Coefficients

For the sake of convenience, the coefficients ⁿC_o, ⁿC₁, ..., ⁿC_r, ..., ⁿC_n are usually denoted by C_o, C₁,..., C_r, ..., C_n respectively

Put x = 1 in (A) and get, 2n = C_o + C₁ + ... + C_n                       ... (D)

Also putting x = -1 in (A) we get,

0 = C_o - C₁ + C₂ - C₃ + ......

=> C₀ + C₂ + C₄ + ...... + C₁ + C₃ + C₅ +...... = 2ⁿ.

Hence C_o + C₂ + C₄ +...... = C₁ + C₃ + C₅ +...... = 2^n-1.

Illustration:

If (1 + x)ⁿ = C₀ + C₁x + C₂x² +...+ C_nxⁿ, then prove that C₀ + (C_o + C₁) + (C₀ + C₁ + C₂) + ... (C₀ + C₁ + C₂ +...+ C_n-1) = n2^n-1 (where n is an even integer.

Solution:

        C₀ + (C₀ + C₁) + (C₀ + C₁ + C₂) +... (C₀ + C₁ + C₂ +...+ C_n-1)

        = C₀+(C₀ + C₁ + C₂ +...+ C_n-1)+(C₀ + C₁)+(C₀ + C₁ + C₂ +...+ C_n-2)+...

        = (C₀ + C₁ + C₂ +...+ C_n)+ (C₀ + C₁ + C₂ +...+ C_n)+... n/2 times

        = (n/2)2ⁿ = n . 2^n-1