Consider the binomial expansion of `(sqrt(x)+(1/(2x^(1/4))))^n n in NN`, where the terms of the expansion are written in decreasing powers of x. If the coefficients of the first three terms form an arithmetic progression then the statement(s) which hold good is(are) (A) total number of terms in the expansion of the binomial is 8 (B) number of terms in the expansion with integral power of x is 3 (C) there is no term in the expansion which is independent of x (D) fourth and fifth are the middle terms of the expansion

Options

A. Total number of terms in the expansion of the binomial is 8
B. Number of terms in the expansion with integral power of x is 3
C. There is no term in the expansion which in indepandent of x
D. Fourth and fifth are the middle terms of the expansion

Answer

b,c

Explanation

` (sqrt(x) + (1)/(2. root (4)(x)))^(n) or (x^(1//2) + (1)/(2) x^(-1//4))^(n)`
` = ""^(n)C_(0) . x^(n/2)+ ""^(n)C_(1) . ((1)/(2)) .x^((2n-3)/(4)) ""^(n)C_(2) . ((1)/(2))^(2) . x^((n-2)/(2) + ...`
Accoding to the question,
` ""^(n)C_(0) ""^(n)C_(1) ((1)/(2)), ""^(n)C_(2) ((1)/(2))^(2) ` are in AP .
` therefore ""^(n)C_(1) = ""^(n)C_(0) + ""^(n)C_(2) ((1)/(2))^(2)`
` rArr n = 1 + (n(n-1))/(4.2)`
` rArr n^(2) - 9n + 8= 0 `
` rArr (n -8) (n-1) = 0 `
` therefore n = 8 ,n ne 1`
option (a) Number of terms = 8 + 1= 9
option (b) Now , ` T_(r+1) = ""^(8)C__(r) . x ^(4 - (r)/(2)). ((1)/(2))^(r) . x^(-(r)/(4))`
` because - le r le 8 `
For integer powers of x, r = 0 , 4, 8
` therefore ` Number of terms in the expansion with integral power of x is 3 .
option (c) Form option (b)
` T_(r +1) = ""^(8)C_(r) . x ^(4 - (3r)/(4)) . ((1)/(2))^(r)`
For indepandent of x
` 4 - (3r)/(4) = 0 `
` r = (16)/(3) cancelin W `
` therefore ` No terms in the given expansion which is indepandent of x
option (d) Middle term is
` T_(s) = ""^(8)C_(4) .x. ((1)/(2))^(4)`
i.e. only one middle term .