# How to find Middle Terms in Binomial Expansion ?

### Middle Terms in Binomial Expansion:

When n is even

Middle term of the expansion is , $\Large (\frac{n}{2} + 1)^{th} term$

When n is odd

In this case $\large (\frac{n+1}{2})^{th} term$ term and$\large (\frac{n+3}{2})^{th} term$ are the middle terms.

e.g. Middle term in the expansion of (1 + x)4 and (1 + x)5

Expansion of (1 + x)4 has 5 terms, so third term is the middle term .

Expansion of (1 + x)5 has 6 terms, so 3rd and 4th both are middle terms .

Illustration : Find the middle term in the expression of

(1 − 2x + x2)n

Solution : (1− 2x + x2)n = [ ( 1 − x)2]n = ( 1− x)2n

Here 2n is even integer, therefore, (2n/2 + 1) th

i.e. (n + 1)th term will be the middle term.

Now (n + 1)th term in (1 − x)2n

= 2nCn (1)2n−n(−x)n = 2nCn(−x)n

$\large = \frac{2n!}{n! n!}(-x)^n$

## Greatest Binomial Coefficient :

Let the (r + 1)th term contains the greatest binomial co-efficient, then

$\large \frac{T_{r+1}}{T_r} = \frac{C_r}{C_{r+1}}= \frac{n-r+1}{r}$

$\large = \frac{n+1}{r}-1 > 1$

$\large \frac{n+1}{r} > 2$

$\large \frac{n+1}{2} > r$ …(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 if n be odd, the greatest binomial coefficient is given when,

$\large r = \frac{n-1}{2} \; or \; \frac{n+1}{2}$

And the coefficients will be $\large n_C{_{(n+1)/2}}$ and $\large n_C{_{(n-1)/2}}$ both being equal .

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

Illustration : Show that the greatest binomial co-efficient in the expansion of :

$\large (x+\frac{1}{x})^{2n}$ is $\large \frac{1.3.5….(2n-1).2^n}{n!}$

Solution: Since middle term has the greatest coefficient,

So, greatest coefficient = coefficient of middle term

= 2nCn

$\large = \frac{1.3.5….(2n-1).2^n}{n!}$

### Greatest Term:

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

$\large |\frac{T_{r+1}}{T_r}|= \frac{n_{C_r} a^{n-r} x^r}{n_{C_{r-1}} a^{n-r+1} x^{r-1}}$

$\large = |\frac{n_{C_r}}{n_{C_{r-1}}}||\frac{x}{a}|$

$\large = |\frac{n-r+1}{r}||\frac{x}{a}|$

$\large = |\frac{n+1}{r}-1 ||\frac{x}{a}|$

Thus |Tr + 1| > |Tr| if

$\large (\frac{n+1}{r}-1 )|\frac{x}{a}| > 1$

Note:

∎ $\large (\frac{n+1}{r}-1 )|$ must be positive since n > r

ThusTr+1 will be the greatest term if, r has the greatest value consistent with the inequality

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

Solution:

$\large \frac{T_{r+1}}{T_r} = \frac{n-r+1}{r}(\frac{3x}{2})$

$\large = \frac{10-r}{r}(\frac{3x}{2})$ ;when x=3/2

$\large = \frac{10-r}{r}(\frac{3}{2})(\frac{3}{2})$

$\large = \frac{10-r}{r}(\frac{9}{4})$

$\large \frac{T_{r+1}}{T_r} = \frac{90-9r}{4r}$

Therefore Tr+1 ≥ Tr if,

90 – 9r ≥ 4r => 90 ≥ 13r

r ≤ 90/13 , r being an integer, hence r = 6

Tr+1 = T7 = T6+1 = 9C6 (2)3 (3x)6

##### Problems based on Greatest Term:

Illustration . Find the greatest term in the expansion of

$\large \sqrt{3}(1 + \frac{1}{\sqrt{3}})^{20}$

Solution: Let rth term be the greatest term.

Since , $\large \frac{T_r}{T_{r+1}} = \frac{r}{21-r}\times \sqrt{3}$

Now , $\large \frac{T_r}{T_{r+1}} \ge 1$

$\large r \ge \frac{21}{\sqrt{3}+1}$ . . . . (1)

Now again ,

$\large \frac{T_{r-1}}{T_r} = \frac{r-1}{22-r}\times \sqrt{3} \le 1$

$\large r \le \frac{22 + \sqrt{3}}{\sqrt{3}+1}$ . . (2)

From (1) and (2) follows that

$\large \frac{21}{\sqrt{3}+1} \le r \le \frac{22 + \sqrt{3}}{\sqrt{3}+1}$

=> r = 8 is the greatest term and its value

$\displaystyle = 20_{C_7}.\frac{1}{27}$

### Properties of Binomial Expansion

∎ There are (n + 1) terms in the expansion of (a + b)n , the first and the last term being an and bn respectively.

∎ $\large \frac{T_{r+1}}{T_r}= \frac{n-r+1}{r} \frac{x}{a}$   for the binomial expansion (a + x)n

∎ Dm (ax + b)n = 0   if m > n where Dm is mth derivative w. r. t. x .

∎ Dm (ax + b)n = ann!   if n = m

∎ Dm (ax + b)n $\large = \frac{a^m \; n!}{(n-m)!}(ax+b)^{n-m}$ ; (m < n)