Question

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

Answer 1

Answer:

tenth term in the expansion  = 1760 / x³

Step-by-step explanation:

(2x²  + 1/x)¹²

10th term in expansion would be  

¹²C₁₀₋₁ (2x²)¹²⁺¹⁻¹⁰(1/x)¹⁰⁻¹

=  ¹²C₉ (2x²)³(1/x)⁹

= {12 * 11 * 10 / (3 * 2 * 1) }  * 2³  * x⁶ / x⁹

= 2 * 11 * 10 * 8 /x³

= 1760 / x³

tenth term in the expansion  = 1760 / x³

Answer 2

Answer:

Step-by-step explanation:

In this question,

We need to find the tenth term in the expansion,

$(2x^{2}\ +\  \frac{1}{x})^{12}$

10th term in expansion will be  

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

=  $^{12}C_{9}(2x^{2})^{3}(\frac{1}{x})^{9}$

= $\frac{12\times 11\times 10}{3\times 2\times 1}\frac{\times2^{3}\times x^{6}}{x^{9}}$

= $\frac{2\times 11\times 10\times 8}{x^{3}}$

= $\frac{1760}{x^{3}}$

Thereofre,Tenth term in the expansion  = $\mathbf{\frac{1760}{x^{3}}}$