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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, Tr+1 > Tr

=>     ((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 nCn/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

        = 2nCn = (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 Tr+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:

         Tr+1/Tr = ((9-r+1)/r) . (2x/3) >1

        i.e. 20 > 5r

If r = 4, then Tr+1 = Tr and these are the greatest terms. Thus 4th and 5th 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  Tr+1 /Tr = ((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 Tr+1 > Tr, if 90 - 9r > 4r.

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

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

Illustration:

Given that the 4th 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 4th 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 rth term be the greatest

Since Tr/Tr+1 = (r/(21-r)).√3.

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

Now again Tr/Tr+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 an-1 x + nC2 an-2 xr +...+ xn.            ...... (1)

(i) Putting x = -x in (1), we get

        (a-x)n = an - nC1 an-1x + nC2 an-2 x2 - nC2 an-3 x3 +...+ (-1)r nCr an-r xr +...+ (-1)n xn.

(ii) Putting a = 1 in (1), we get,

        (1+x)n = nC0 + nC1x + nC2x2 +...+ nCr xr +...+ nCnxn.                    ... (A)

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

        (1-x)n=nC0-nC1 x + nC2 x2-nC3x3 +... (-1)r nCr xr +... (-1)n nCnxn     ... (B)

Tips to Remember

(a) Tr/Tr+1 = ((n-r+1)/r ).(x/a) for the binomial expansion of (a + x)n.

(b) (n+1)Cr = nCr + nCr-1.

(c) r nCr = n n-1Cr-1

(d) When n is even,

        (x + a)n + (x - a)n = 2(xn + nC2 xn-2 a2 + nC4 xn-4 a4 +...+ nCn an).

When n is odd,

        (x + a)n + (x - a)n = 2(xn + nC2 xn-2 a2 +...+ nCn-1 x an-1).

When n is even

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

When n is odd

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



Properties of Binomial Coefficients

For the sake of convenience, the coefficients nCo, nC1, ..., nCr, ..., nCn are usually denoted by Co, C1,..., Cr, ..., Cn respectively

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

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

0 = Co - C1 + C2 - C3 + ......

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

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

Illustration:

If (1 + x)n = C0 + C1x + C2x2 +...+ Cnxn, then prove that C0 + (Co + C1) + (C0 + C1 + C2) + ... (C0 + C1 + C2 +...+ Cn-1) = n2n-1 (where n is an even integer.

Solution:

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

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

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

        = (n/2)2n = n . 2n-1

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