Search any question & find its solution
Question:
Answered & Verified by Expert
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) $O^2-E^2=\left(x^2-a^2\right)^n$.
(ii) $4 O E=(x+a)^{2 n}-(x-a)^{2 n}$.
(i) $O^2-E^2=\left(x^2-a^2\right)^n$.
(ii) $4 O E=(x+a)^{2 n}-(x-a)^{2 n}$.
Solution:
2831 Upvotes
Verified Answer
(i) Given expression is $(x+a)^n$.
$$
\begin{aligned}
&\therefore(x+a)^n={ }^n C_0 x^n a^0+{ }^n C_1 x^{n-1} a^1+{ }^n C_2 \\
&x^{n-2} a^2+{ }^n C_3 x^{n-3} a^3+\ldots+{ }^n C_n a^n
\end{aligned}
$$
Now, sum of odd terms
$$
O={ }^n C_0 x^n+{ }^n C_2 x^{n-2} a^2+\ldots
$$
and sum of even terms $E$
$={ }^n C_1 x^{n-1} a+{ }^n C_3 x^{n-3} a^3+\ldots$
$\because \quad(x+a)^n=O+E$
Similarly, $(x-a)^n=O-E$
multiply eqs. (i) and (ii)
we have $(O+E)(O-E)=(x+a)^n(x-a)^n$
$\Rightarrow O^2-E^2=\left(x^2-a^2\right)^n$
(ii) $4 O E=(O+E)^2-(O-E)^2$
$=\left[(x+a)^n\right]^2-\left[(x-a)^n\right]^2$ [using eqs. (i) and (ii) $]$
$=(x+a)^{2 n}-(x-a)^{2 n}$
$$
\begin{aligned}
&\therefore(x+a)^n={ }^n C_0 x^n a^0+{ }^n C_1 x^{n-1} a^1+{ }^n C_2 \\
&x^{n-2} a^2+{ }^n C_3 x^{n-3} a^3+\ldots+{ }^n C_n a^n
\end{aligned}
$$
Now, sum of odd terms
$$
O={ }^n C_0 x^n+{ }^n C_2 x^{n-2} a^2+\ldots
$$
and sum of even terms $E$
$={ }^n C_1 x^{n-1} a+{ }^n C_3 x^{n-3} a^3+\ldots$
$\because \quad(x+a)^n=O+E$
Similarly, $(x-a)^n=O-E$
multiply eqs. (i) and (ii)
we have $(O+E)(O-E)=(x+a)^n(x-a)^n$
$\Rightarrow O^2-E^2=\left(x^2-a^2\right)^n$
(ii) $4 O E=(O+E)^2-(O-E)^2$
$=\left[(x+a)^n\right]^2-\left[(x-a)^n\right]^2$ [using eqs. (i) and (ii) $]$
$=(x+a)^{2 n}-(x-a)^{2 n}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.