Question

In the binomial expansion of
(k+1/2x)^9
where k is a positive constant
the coefficient of x³ is 70
Find the value of k, giving your answer to 2 decimal places.

Answer 1

Answer:

k = 1.37

Step-by-step explanation:

We can find the coefficient of x³ in the expansion of $\bf (k+\frac{1}{2} x)^9$ by using the Binomial Formula:

$\boxed{_nC_i\cdot x^{n-i}\cdot y^i}$

Given:

• $n=9$ (highest degree of polynomial = 9)
• $n-i=3$ (degree of x³ = 3)

        $9-i=3$

              $i=6$

• $x=\frac{1}{2} x$
• $y=k$

The coefficient of x³ $=_9C_6\cdot (\frac{1}{2} x)^{9-6}\cdot (k)^6$

                          $\displaystyle70x^3=\frac{9!}{6!3!} \cdot\left(\frac{1}{2} x\right)^3\cdot k^6$

                    $\displaystyle70x^3=\frac{9\times8\times7}{3\times2\times1} \left(\frac{1}{8}x^3 \right)(k)^6$

                        $\displaystyle70=\frac{21}{2} k^6$

                         $\displaystyle k=\sqrt[6]{\frac{70\times2}{21} }$

                         $\bf k=1.37$