Mathematics General Term , Middle Terms and Constant term
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### 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.