Mathematics General Term , Middle Terms and Constant term

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Class 11 Chapter 8 - BINOMIAL THEOREM

General Term , Middle Terms and Constant term

Topic Covered

`star` General Term
`star` Middle Terms
`star` Constant term

General Term

`\color{green} ✍️` In the binomial expansion for

`color{blue} ((a + b)^n = ""^nC_0a^n + ""^nC_1a^(n–1)b + ""^nC_2a^(n–2) b^2 + ......+ ""^nC_(n – 1)a.b^(n–1) + ""^nC_nb_n)`

`color{green} ✍️` we observe that the first term is `""^nC_0a^n,` the second term is `""^nC_1a^(n–1)b,` the third term is `""^nC_2a^(n–2)b^2,` and so on. Looking at the pattern of the successive terms we can say that the `(r + 1)^(th)` term is `""^nC_ra^(n–r)b^r.`

`\color{green} ✍️` The `(r + 1)^(th)` term is called the `\color{green}("general term")` of the expansion `(a + b)^n.` It is denoted by `T_(r+1).`

Thus `T_(r+1) = ""^nC_ra^(n–r)b^r.`

Middle Terms

Regarding the middle term in the expansion `(a + b)^n` , we have

`(i)` If `color{green}("n is even")`, then the number of terms in the expansion will be `n + 1.`

Since `n` is even so `n + 1` is odd. Therefore, the middle term is `((n+1+1)/2)^(th)` i.e., ` (n/2+1)^(th)` term.

`(ii)` If `color{green}("n is odd")` , then `n +1` is even, so there will be two middle terms in the expansion, namely, `((n+1)/2)^(th)` term and `((n+1)/2+1)^(th)` term.

Constant term

`\color{green} ✍️` In the expansion of `(x+1/x)^(2n)` where `x ≠ 0,` the middle term is `((2n+1+1)/2)^(th)` i.e., `(n + 1)^(th)` term, as `2n` is even.

It is given by `""^(2n)C_nx^n(1/x)^n = ""^(2n)C_n` (constant).

`\color{green} ✍️` This term is called the term `color{green}("independent of x")` or the constant term.

Q 3121556421

Find a if the 17th and 18th terms of the expansion `(2 + a)^(50)` are equal.

Solution:

The `(r + 1)th` term of the expansion `(x + y)^n` is given by `T_(r + 1) = nC_r x^(n–r)y^r.` For the 17th term, we have, `r + 1 = 17`, i.e., `r = 16`

Therefore, `T_(17) = T_(16+1) = text()^(50)C_(16) (2)^(50-16) a^(16)`

`= text()^(50)C_(16) 2^(34) a^(16)`

Similarly, `T_(18) = text()^(50)C_(17) 2^(33) a^(17)`

Given that `T_(17) = T_(18)`

So `text()^(50)C_(16) (2)^(34) a^(16) = text()^(50)C_(17) (2)^(33) a^(17)`

Therefore `{ text()^(50)C_(16) . 2^(34) }/{ text()^(50)C_(17) . 2^(33)} = a^(17)/a^(16)`

`a = { text()^(50)C_(16) xx 2}/(text()^(50)C_(17)) = (50!)/(16 ! 34!) xx (17 ! . 33!)/(50!) xx2 = 1`

Q 3111556429

Show that the middle term in the expansion of `(1+x)^(2n)` is `( 1.3.5..... (2n-1))/(n!) 2n x^n` where n is a positive integer.

Solution:

As `2n` is even, the middle term of the expansion `(1+x)^(2n)` is `((2n)/2+1)^(th)` i.e., `(n + 1)th` term which is given by,

`T_(n+1) = text()^(2n) C_1 (1)^(2n-n) (x)^n = text()^(2n)C_n x^n = {(2n)!}/(n! n!) x^n`

` = { 2n (2n-1) (2n-2) .........4.3.2.1}/(n! n!) x^n`

` = { 1.2.3.4........(2n-2)(2n-1)(2n)}/(n! n!) x^n`

` = {[ 1.3.5........(2n-1) ] [2.4.6 ........(2n) ]}/(n! n!) x^n`

` = { [1.3.5 ....(2n-1) ]2^n [ 1.2.3.....n]}/(n! n! ) x^n`

` = {[1.3.5.......(2n-1) ] n! }/(n! n! ) 2^n . x^n`

` = {1.3.5 ......(2n-1)}/(n! ) 2^n x^n`

Q 3131656522

Find the coefficient of `x^6 y^3` in the expansion of `(x+2y)^9`

Solution:

Suppose `x^6y^3` occurs in the `(r + 1)th` term of the expansion `(x + 2y)^9.`

Now `T_(r+1) = text()^9C_r x^(9-r) (2y)^r = text()^9C_r 2^r . x^(9-r) . y^r`

Comparing the indices of `x` as well as `y` in `x^6y^3` and in `T_(r + 1) `, we get `r = 3.`
Thus, the coefficient of `x^6y^3` is `text()^9C_3 2^3 = (9!)/(3! 6!) .2^3 = (9.8.7)/(3.2) .2^3 = 672`

Q 3131156922

The coefficients of three consecutive terms in the expansion of `(1 + a)^n` are in the ratio `1: 7 : 42`. Find `n`.

Solution:

Suppose the three consecutive terms in the expansion of `(1 + a)^n` are `(r – 1)th`, `rth` and `(r + 1)th` terms.
The `(r – 1)th` term is `text()^nC_(r – 2) a_(r – 2)`, and its coefficient is `text()^nC_(r – 2)`. Similarly, the coefficients of `rth` and `(r + 1)th` terms are `text()^nC_(r – 1)` and `text()^nC_r` , respectively. Since the coefficients are in the ratio `1 : 7 : 42`, so we have,

`{ text()^nC_(r-2)}/{text()^nC_(r-1)} = 1/7 , i.e. n-8r+9 = 0` .............(1)

and `{ text()^nC_(r-1)}/(text()^nC_r) = 7/42 , i.e. n-7r+1 = 0` ............(2)

Solving equations(1) and (2), we get, `n = 55.`