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

$=t_1+t_2+t_3+\cdots +t_{r+1}\cdots t_n$

Given that

${\left(x-a\right)}^n={}^n{\text{C}}_0{x}^n-{}^n{\text{C}}_1.x^{n-1}.a+{}^n{\text{C}}_2{x}^{n-2}.a^2+\cdots +{\left(-1\right)}^r{}^n{\text{C}}_r{x}^{n-r}.a^r+\cdots +{\left(-1\right)}^n{}^n{\text{C}}_n.a^n$ ${\left(x+a\right)}^n+{\left(x-a\right)}^n=2\left[t_1+t_3+t_5+\cdots \right]=p$

$\frac{{\left(x+a\right)}^n+{\left(x-a\right)}^n}{2}=p\cdots \cdots \left(1\right)$

Simplify ${\left(x+a\right)}^n-{\left(x-a\right)}^n=2\left[t_2+t_4+t_6+\cdots \right]=q$

$\frac{{\left(x+a\right)}^n-{\left(x-a\right)}^n}{2}=q\cdots \cdots \cdots \left(2\right)$

Multiply $\left(1\right)\&\left(2\right)$

$\left[\frac{{\left(x+a\right)}^n+{\left(x-a\right)}^n}{2}\right]\left[\frac{{\left(x+a\right)}^n-{\left(x-a\right)}^n}{2}\right]=pq$

${\left(x+a\right)}^{2n}-{\left(x-a\right)}^{2n}=4pq$