Question

Question 8 Find the middle terms in the expansions of (x/3 + 9y)^10

Class X1 - Maths -Binomial Theorem Page 171

Answer 1

(x/3 + 9y)^10
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 = 10
total number of terms = n + 1 = 10 + 1 = 11 { odd}
so, middle term = { (n+1 +1)/2}th term
= { (10+1+1)/2}th term
= 6th term

now, The general term in the expansion of (x/3 + 9y)^10 is
T_{r+1} = 10Cr.(x/3)^{10-r}(9y)^r [ by formula of the general term , T _{r+1}=nCr.x^{n-r}y^r]

putting r = 5 ,
T_{5+1} = T_6 = 10C5.(x/3)^{10-5}(9y)^5
= {10!/5!×5!} (x/3)^5 (9)^5 y^5
= {10 × 9 × 8 × 7 × 6/5 × 4 × 3 × 2}×9^5 × x^5y^5/3^5
= {9 × 8 × 7 × 2/4 } × x^5y^5 × 3^5
= 9 × 7 × 4 × 243 × x^5y^5
= 61236x^5y^5