My only way to express sorry

First we note coefficient of $x^r$ in expansion of $(1+x)^n$ is $\binom{n}{r}$ .
According to the question: $\large \binom{15}{2r}>\frac{1}{2}\binom{15}{2r+1}$ $\implies \dfrac{\not{15!}}{(15-2r)!(2r)!}>\dfrac{\not{15!}}{2(15-2r-1)!(2r+1)!}$ $\implies \dfrac{1}{15-2r}>\dfrac{1}{2(2r+1)}$ $\implies \dfrac{6r-13}{2r-15}<0$ $\implies \large \frac{13}{6}<r<\frac{15}{2}$ Hence integral r=3,4,5,6,7.
$\therefore~3+4+5+6+7=\huge\color{#456461}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{ 25}}}}}$