Question

Independent of x in the expansion of ?

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

Answer 1

`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`

Answer 2

`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.!