# In the expansion of $(x+a)^n$ if the sum of odd terms is denoted by O and the sum of even term by E.Then prove that (i) $O^2-E^2=(x^2-a^2)^n$

Toolbox:
• $(x+a)^n=nC_0x^na^0+nC_1x^{n-1}a^1+nC_2x^{n-2}a^2+nC_3x^{n-3}a^3+.......+nC_n x^0a^n$
$(x+a)^n=nC_0x^na^0+nC_2x^{n-2}a^2+......)+nC_1 x^{n-1}a^1+nC_3x^{n-3}a^3+.....)$
$\therefore (x+a)^n=O+E$------(1)
Also $(x-a)^n =nC_0x^n a^0-nC_1 x^{n-1}a^1+nC_2x^{n-2}a^2-nC_3x^{n-3}a^3+......+(-1)^n nC_nx^0a^n$
$\therefore (x-a)^n=(nC_0x^n a^0+x_2 x^{n-2}a^2+.....)-(nC_1x^{n-1}a^1+nC_3 x^{n-3}a^3)$
$(x-a)^n$=O-E-----(2)
Now $O^2-E^2=(O+E)(O-E)$
$\Rightarrow (x+a)^n (x-a)^n$
$\Rightarrow (x^2-a^2)^n$
Hence proved.