Question

If $(1+x{)}^{2n}=a_0+a_{1}x....+a_{2n}{x}^{2n}$, then

• A. $a_1+a_2+a_4.....=\frac{1}{2}(a_0+a_1+a_2.....)$
• B. $a_{n+1}=a_n$
• C. $a_{n-3}=a_{n+3}$
• D. $a_{n-3}>a_{n+3}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a_1+a_2+a_4.....=\frac{1}{2}(a_0+a_1+a_2.....)$
B $a_{n+1}=a_n$
C $a_{n-3}=a_{n+3}$

${(1+x)}^{2n}+{(1-x)}^{2n}=2(a_0+a_2{x}^2+a_4{x}^4+...+a_{2n}{x}^{2n})$

For $x=1$, $\frac{2^{2n}}{2}=(a_0+a_2+a_4+...+a_{2n})$

${(1+x)}^{2n}=(a_0+a_{1}x+a_2{x}^2+...+a_{2n}{x}^{2n})$ ...$\left(i\right)$

For $x=1$, $2^{2n}=a_0+a_1+a_2+....+a_{2n}$

Hence,

$(a_0+a_2+a_4+...+a_{2n})=$ $1/2(a_0+a_1+a_2+....+a_{2n})$

So, $A$ is correct.

Now, $\frac{(1+x{)}^{2n}}{x^{2n}}=\frac{(a_0+a_{1}x+...+a_{2n}{x}^{2n})}{x^{2n}}$

So, ${(1+\frac{1}{x})}^{2n}=\frac{a_o}{x^{2n}}+\frac{a_1}{x^{2n-1}}+....+\frac{a_{2n}}{x^0}$ ...$\left(ii\right)$

Putting $x=\frac{1}{x}$ in $\left(i\right)$,

${(1+\frac{1}{x})}^{2n}=a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+...+\frac{a_{2n}}{x^{2n}}$ ...$\left(iii\right)$

On comparing coefficients from $\left(ii\right)$ and $\left(iii\right)$, $a_{2n}=a_0$, $a_{2n-1}=a_1$.... and so on...

So, we can say, $a_{2n-r}=a_r$. for $2n\ge r\ge 0$

For, $r=n+3$, $a_{n-3}=a_{n+3}$ So, $C$ is correct.

But for $r=n$, we find $B$ is incorrect.