Algebra + Calculus = Algebrus

We know that ${ U }_{ n }=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1-\cos (2nx) }{ 1-\cos (2x) } } dx\\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 2{ sin }^{ 2 }nx }{ 2{ sin }^{ 2 }x } } dx\\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sin }^{ 2 }nx }{ { sin }^{ 2 }x } } dx\\ =\frac { n\pi }{ 2 }$ Hence, the value of $\triangle$ can be written as $\left| \begin{matrix} \frac { \pi }{ 2 } & \pi & \frac { 3\pi }{ 2 } \\ 2\pi & \frac { 5\pi }{ 2 } & 3\pi \\ \frac { 7\pi }{ 2 } & 4\pi & \frac { 9\pi }{ 2 } \end{matrix} \right|$ ${ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 },{ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\\ =\left| \begin{matrix} \frac { \pi }{ 2 } & \frac { \pi }{ 2 } & \pi \\ 2\pi & \frac { \pi }{ 2 } & \pi \\ \frac { 7\pi }{ 2 } & \frac { \pi }{ 2 } & \pi \end{matrix} \right|$ $\\ =\frac { { \pi }^{ 2 } }{ 2 } \left| \begin{matrix} \frac { \pi }{ 2 } & 1 & 1 \\ 2\pi & 1 & 1 \\ \frac { 7\pi }{ 2 } & 1 & 1 \end{matrix} \right| =0$

${ U }_{ n }+{ U }_{ n+2 }-{ 2U }_{ n+1 }=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \left( 1-cos2nx \right) +\left[ 1-cos\left( 2n+4 \right) x \right] -2\left[ 1-cos\left( 2n+2 \right) x \right] }{ (1-cos2x) } } dx$

$=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { -2cos\left( \left( 2n+2 \right) x \right) .cos2x+2cos\left( \left( 2n+2 \right) x \right) }{ (1-cos2x) } } dx$

$=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ 2cos\left( \left( 2n+2 \right) x \right) } dx$

$=2{ \left[ \frac { sin\left( 2n+2 \right) x }{ 2n+2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=0$

$\therefore \quad \quad { U }_{ n }+{ U }_{ n+2 }-{ 2U }_{ n+1 } \ = \ 0 \quad \quad \quad \quad \quad ...(1)$

Now applying, $C_{ 1 }+{ C }_{ 3 }-2C_{ 2 }$ , then by relation (1) each element in new ${C}_{1}$ will be zero corresponding to n=1, 4, 7 .

$\therefore \quad \quad \quad \left\lfloor \triangle \right\rfloor =\left\lfloor 0 \right\rfloor =\boxed{0}$