Independent of x in the expansion of ?

#(4x^3+1/2x)^8#

2 Answers
Aug 13, 2017

#7#

Explanation:

Assuming that the formula is #(4 x^3 + 1/(2 x))^8#

#(X+Y)^n = sum_(k=1)^n((k),(n))X^k Y^(n-k)#

Here #X=4x^3# and #Y = (2x)^-1# so we must solve

#((4x^3))^k((2x)^-1)^(8-k)=4^k2^(k-8)x^(3k+k-8)#

for #3k+k-8=0# giving #k = 2# and then

#a_0=((2),(8))xx4^2xx2^-6 = (8!)/((2!)(6!))2^-2 = 7#

# 7.#

Explanation:

I hope, the Question is to find a term independent of #x# in

the Expansion of #(4x^3+1/(2x))^8.#

In the Expansion of #(a+b)^n#, the General #(r+1)^(th)# Term,

denoted by, #T_(r+1),# nd it is is given by,

#T_(r+1)=""_nC_ra^(n-r)b^r, r=0,1,2,...,n.#

In our Problem, we have, #a=4x^3, b=1/(2x), n=8,# so that,

#T_(r+1)=""_8C_r(4x^3)^(8-r)*(1/(2x))^r,#

#=""_8C_r(2^2)^(8-r)x^(24-3r)2^-rx^-r,#

# rArr T_(r+1)=""_8C_r*2^(16-3r)*x^(24-4r), r=0,1,...,8.#

For the term independent of #x,# we must have, #24-4r=0.#

#:. r=6.#

#:." the Reqd. Term="T_7=""_8C_6*2^(16-18),#

#=""_8C_22^-2={(8*7)/(1*2)}(1/4)=7.#

Enjoy Maths.!