Greatest Binomial Coefficient

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 nC_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 nC_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) 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

Free study material for the preparation of IIT JEE, AIEEE and other engineering examinations is available online at askIITians.com. Study Physics, Chemistry and Mathematics at askIITians website and be a winner. We offer numerous live online courses as well for live online IIT JEE preparation - you do not need to travel anywhere any longer - just sit at your home and study for IIT JEE live online with askIITians.com

To read more, Buy study materials of Binomial Theorem comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.