Question

Question 7 Find the middle terms in the expansions of (3 - x^3 / 6) ^ 7

Class X1 - Maths -Binomial Theorem Page 171

Answer 1

(3 - x³/6)^7

concept :-
suppose total number of terms is (n +1) ,e.g odd number . then middle term is {(n+1 + 1)/2} th term and when total number of terms is even , then the middle terms are {(n+1)/2}th. and {(n+3)/2}th term .

solution :- Here, n= 7
total number of terms = n +1 = 7 + 1 = 8
middle term is obtained by {(n+1)/2}th and {(n+3)/2}th term .
e.g {(7+1)/2}th term and {(7+3)/2}th term
=> 4th term and 5th term

now,
T_{r+1} =nCr.x^{n-r}(-y)^r [ by formula of general term]
T_{3+1} = T_4 = 7C3.(3)^{7-3}(-x^3/6)^3
= -{7 × 6 × 5/6} × { 3⁴ × x^9/216}
= - 35 × 27 × x^9/72
= -105x^9/8

and
T_{4+1} = T_{5} = 7C4.(3)^{7-4}(-x^3/6)^4
= {7 × 6 × 5/6} (3)³ × {(-1)⁴ × x¹²/6⁴ }
= 35 × 27 × x¹²/216 × 6
= 35x¹²/48