Question

Find the constant term in the expansion of
$\rm \Big(2x^{2} - \frac{1}{x} \Big)^{12}$

Answer 1

hn....here..this is the crrct solution...maybe calculation could get wrong...but this is the right way

Answer 2

Answer:

16$C_{8}$

Step-by-step explanation:

We have to find the constant term in the expansion of $(2x^{2}-\frac{1}{x}  )^{12}$. ........ (1)

We know the following binomial expansion

$(a+x)^{n}=a^{n} +C_{1} a^{n-1}x + C_{2} a^{n-2}x^{2} + ....... +C_{n}x^{n}$ ....... (2)

Here from (1), the binomial expansion will be

$(2x^{2}-\frac{1}{x}  )^{12}= (2x^{2} )^{12} +C_{1} (2x^{2} )^{11}(-\frac{1}{x} )+C_{2} (2x^{2} )^{10}(-\frac{1}{x} )^{2}+ ....... +(-\frac{1}{x} )^{12}$

= $2^{12}.x^{24} -C_{1}.2^{11}.x^{21}+C_{2}.2^{10}.x^{18}-C_{3}.2^{9}.x^{15}+ ...... +C_{n}\frac{1}{x^{12} }$

Hence, in the above expansion, the power of x decreases from 24 in the first term to 21 in the second term, then again from 21 in the second term to 18 in the third term and so on.

So, the power of x will reduce to 0 in the ninth term and it will be

$t_{9}=C_{8}.(2x^{2} )^{4}.(-\frac{1}{x} )^{8}$

              =16$C_{8}$ ... This is the constant term. (Answer)