Question

if in the expansion of (1+x)^n the coefficient of pth and qth terms are equal prove that p+q=n+2 where p is not equal to q​

Answer 1

We know,

$\displaystyle (1+x)^n=\sum_{k=0}^{n}\ ^n\!C\!_k\ x^k$

So we can see that coefficient of $x^i$ is $^n\!C_i$ for $0\leq i\leq n.$

Hence the coefficients of $x^p$ and $x^q$ are respectively $^n\!C_p\quad\&\quad\ ^n\!C_q.$

But, remember, the first term in the expansion is $^n\!C_0x^0.$ So $^n\!C_ix^i$ is the $(i+1)^{th}$ term of the expansion.

So the $p^{th}$ and $q^{th}$ terms are respectively $^n\!C_{p-1}\quad\&\quad\ ^n\!C_{q-1}.$

Given, $^n\!C_{p-1}=\ ^n\!C_{q-1}$

Since $p\neq q,$ we can remember $^n\!C_r=\ ^n\!C_{n-r}$

So we can say,

$^n\!C_{n-(p-1)}=\ ^n\!C_{q-1}\\\\\implies\ \ n-p+1=q-1\\\\\implies\ \ p+q=n+2$

Hence Proved!