If the sum of odd terms and sum of even terms in ${(\mathrm{x} + \mathrm{a})}^{\mathrm{n}}$ are p and q respectively, then 4pq =

$\begin{array}{rcl}\mathrm{P}& =& {}^{\mathrm{n}}{\mathrm{C}}_0 {\left(\mathrm{x}\right)}^{\mathrm{n}}+{}^{\mathrm{n}}{\mathrm{C}}_2 {\left(\mathrm{a}\right)}^{\mathrm{n}-2} {\mathrm{a}}^2+.....\\ \mathrm{q}& =& {}^{\mathrm{n}}{\mathrm{C}}_1 {\left(\mathrm{x}\right)}^{\mathrm{n}-1} \mathrm{a}+{}^{\mathrm{n}}{\mathrm{C}}_3 {\mathrm{x}}^{\mathrm{n}-3} {\mathrm{a}}^3+......\\ {(\mathrm{x}+\mathrm{a})}^{\mathrm{n}}& =& \mathrm{p}+\mathrm{q}, {(\mathrm{x}-\mathrm{a})}^{\mathrm{n}}=(\mathrm{p}-\mathrm{q})\\ {(\mathrm{x}+\mathrm{a})}^{2\mathrm{n}}& =& {(\mathrm{p}+\mathrm{q})}^2, {(\mathrm{x}-\mathrm{a})}^{2\mathrm{n}}={(\mathrm{p}-\mathrm{q})}^2 \\ 4\mathrm{pq}& =& {(\mathrm{p}+\mathrm{q})}^2-{(\mathrm{p}-\mathrm{q})}^2={(\mathrm{x}+\mathrm{a})}^{2\mathrm{n}}-{(\mathrm{x}-\mathrm{a})}^{2\mathrm{n}}\end{array}$