Question

Find the 10th term in the binomial expansion of (2x^2+1/x)^12

Answer 1

Answer:

264(x^-6)

Step-by-step explanation:

given in the question an expression

(2x^2 + 1/x)^12

To Find the 10th term of above expression we will use following formula

nCr (2x²)^(n-r) (1/x)^(r)

where n = 12

            r = 10

12C10 (2x²)^2 (1/x)^10

66 (4x^4) (1/x^10)

264(x^4) (1/x^10)

we will apply power product rule  

264(x^(4-10))

The 10th term in the binomial expansion of (2x^2+1/x)^12 = 264(x^-6)

Answer 2

Answer:

$T_{10}=2760\:x^{-3}$

Step-by-step explanation:

Formula used:

The (r+1)th term in the expansion of $(a+b)^n$ is

$T_{r+1}=nC_r\:a^{n-r}\:b^r$

Now,

$(2x^2+\frac{1}{x})^12$

$a=2x^2,b=\frac{1}{x},n=15$

$T_{r+1}=nC_r\:a^{n-r}\:b^r$

$T_{r+1}=12C_r\:(2x^2)^{12-r}\:(\frac{1}{x})^r$

$T_{r+1}=12C_r\:2^{12-r}\:x^{24-2r}\:(x^{-1})^r$

$T_{r+1}=12C_r\:2^{12-r}\:x^{24-2r}\:x^{-r}$

$T_{r+1}=12C_r\:2^{12-r}\:x^{24-3r}$

put n=9 we get

$T_{10}=12C_9\:2^{12-9}x^{24-3(9)}$

$T_{10}=12C_9\:2^{3}x^{-3}$

$T_{10}=\frac{12*11*10}{1*2*3}\:(8)x^{-3}$

$T_{10}=(4*11*5)\:(8)x^{-3}$

$T_{10}=2760\:x^{-3}$