Question

If $\sum _{k=0}^4\left(\frac{3^{4-k}}{(4-k)!}\right)\left(\frac{x^k}{k!}\right)=\frac{32}{3},$ then the largest value of $x$ is

Answer 1

$\sum _{k=0}^4\left(\frac{3^{4-k}}{(4-k)!}\right)\left(\frac{x^k}{k!}\right)=\sum _{k=0}^4\left(\frac{3^{4-k}}{(4-k)!}\right)\left(\frac{x^k}{k!}\right)\frac{4!}{4!}=\sum _{k=0}^4\left(\frac{4!\cdot 3^{4-k}{x}^k}{(4-k)!k!}\right)\frac{1}{4!}=\sum _{k=0}^4\frac{{}^4{C}_k\cdot 3^{4-k}\cdot x^k}{4!}=\frac{(3+x{)}^4}{4!}$
Therefore,
$\frac{(3+x{)}^4}{4!}=\frac{32}{3}\Rightarrow (3+x{)}^4=2^8\Rightarrow (x+3)=\pm 4$
$\Rightarrow x=1\text{or}-7$
So, the largest real value of $x$ is $1.$