Question:
In the expansion of $(x+a)^{n}$ if the sum of odd terms is denoted by 0 and the sum of even term by E. Then prove that
(i) $\mathrm{O}^{2}-\mathrm{E}^{2}=\left(\mathrm{x}^{2}-\mathrm{a}^{2}\right)^{n}$
(ii) $40 E=(x+a)^{2 n}-(x-a)^{2 n}$
Solution:
(i) We know that
$(x+a)^{n}={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} a^{1}+{ }^{n} C_{2} x^{n-2} a^{2}+{ }^{n} C_{3} x^{n-3} a^{3}+\ldots$
Sum of odd terms,
$O={ }^{n} C_{0} x^{n}+{ }^{n} C_{2} x^{n-2} a^{2}+\ldots$
And also sum of even terms
$E={ }^{n} C_{1} x^{n-1} a+{ }^{n} C_{3} x^{n-3} a^{3}+\ldots$
Since $(x+a)^{n}=0+E$
$(x-a)^{n}=0-E$
Therefore,
$(O+E)(O-E)=(x+a)^{n}(x-a)^{n}$
(ii) $40 E=(O+E)^{2}-(O-E)^{2}$
$=(x-a)^{2 n}-(x-a)^{2 n}$