Question

If $a$ is the coefficient of the middle term in the expansion of $(1+x{)}^{2n}$ and $b,c$ are the coefficients of the two middle terms in the expansion of $(1+x{)}^{2n-1}$ then

• A. $a+b=c$
• B. $a=b+c$
• C. $a=b=c$
• D. $b=a+c$

Answer 1

The correct option is B $a=b+c$

For $(1+x){}^{2n}$
Middle term $=\frac{N}{2}+1$ since 2n is even
$=\frac{2n}{2}+1$
$=(n+1{)}^{th}$ term
Now consider the following
$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$...(i)
Where $T$ represents the term
Here the middle term is the $(n+1{)}^{th}$ term
Hence $r+1=n+1$
$r=n$
Substituting in (i), we get
${}^{2n}{C}_n{1}^{2n-n}{x}^n$
$={}^{2n}{C}_n{x}^n$
Therefore $a={}^{2n}{C}_n$ because a is the coefficient ...(a)
Consider $(1+x){}^{2n-1}$
Since $2n-1$ is odd the middle terms are $\left(\frac{N+1}{2}\right)$th and $(\frac{N+1}{2}+1)$ th terms
Now consider the following
$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$...(ii)
Where $T$ represents the term
Here the middle terms are the $(n{)}^{th}$ term and $(n+1{)}^{th}$ term.
Hence $r+1=n+1$ and $r+1=n$
$r=n$ and $r=n-1$
For $r=n$
$T_{r+1}={}^{2n-1}{C}_n{1}^{n-1}{x}^n$
$={}^{2n-1}{C}_n{x}^n$
Hence, $b={}^{2n-1}{C}_n$, since $b$ is the coefficient ...(b)
Similarly for $r=n-1$
$T_{r+1}={}^{2n-1}{C}_{n-1}{1}^n{x}^{n-1}$
$={}^{2n-1}{C}_{n-1}{x}^{n-1}$
Hence $c={}^{2n-1}{C}_{n-1}$ since $c$ is the coefficient ...(c)
From $a,b,c$, we get
${}^{2n}{C}_n={}^{2n-1}{C}_n+{}^{2n-1}{C}_{n-1}$
$a=b+c$