700 followers problem (2 in 1)

The binomial expansion of the expression is:

$\displaystyle \left( 4x^2+\dfrac {1}{2x} \right)^{15} = \sum _{n=0} ^{15} {\begin{pmatrix} 15 \\ n \end{pmatrix} (4x^2)^{15-n} (2x)^{-n}}$

$\displaystyle \quad \quad \quad \quad \quad \quad = \sum _{n=0} ^{15} {\begin{pmatrix} 15 \\ n \end{pmatrix} 2^{2(15-n)-n} x^{2(15-n)-n}} = \sum _{n=0} ^{15} {\begin{pmatrix} 15 \\ n \end{pmatrix} \left( 2x \right) ^{30-3n}}$

The term without $x$ is when $30-3n=0 \quad \Rightarrow n = 10$ .

The term $y = \begin{pmatrix} 15 \\ 10 \end{pmatrix} = 3003$

The middle two terms are when $n=7$ and $n=8$ , and when they are equal, $x=z$ is given by:

$\Rightarrow \begin{pmatrix} 15 \\ 7 \end{pmatrix} (2z)^{30-21} = \begin{pmatrix} 15 \\ 8 \end{pmatrix} (2z)^{30-24} \quad \Rightarrow (2z)^3 = 1\quad \Rightarrow z = \frac{1}{2}$

Therefore, $\quad y+z = \boxed{3003.5}$

we can find y by taking equation 2(15-p)=p where p is the term in binomial expansion hence p=10 hence value of $y=\frac{15!}{10! 5!}=3003$

Now as for z, middle terms are 7 and 8

and we know 15C8=15C7

thus for 8th term $\frac{((2x)^{2})^{7}}{(2x)^{8}} =(2x)^{6}$

and for 7 th term $\frac{((2x)^{2})^{8}}{(2x)^{7}} = (2x)^{9}$

making both equal $(2x)^{9} = (2x)^{6}$

dividing $(2x)^{(9-6)}=1$

$(2x)^{3}=1$

hence z= x=0.5 and y+z=3003.5