Question

Prove that the coefficient of $x^m$ in the expansion of $(1+x{)}^m+2(1+x{)}^{n-1}+3(1+x{)}^{n-2}+(n-m+1\left)\right(1+x{)}^m$ where $0\ge m\ge nis{}^{n+2}{C}_{m+2}$

Answer 1

The given series is an A.G.S with d = 1 and r = $\frac{1}{1+x}$
$S=(1+x{)}^n+2(1+x{)}^{n-1}+3(1+x{)}^{n-2}$
$\therefore$ ....(n - 1 + m) terms
$S\frac{1}{1+x}=(1+x{)}^{n-1}+2(1+x{)}^{n-2}+.......+(n-m+1)(1+x{)}^{m-1}$
Subtracting $S[1-\frac{1}{1+x}]=(1+x{)}^n+(1+x{)}^{n-1}+1+x{)}^{n-2}.....$
or$\frac{x}{1+x}S=(1+x{)}^n\left[\frac{-1{\left(\frac{1}{1+x}\right)}^{n-m+1}}{1-1/(1+x)}\right]$
-(n - m + 1)(1 + x)${}^{m-1}$
$\therefore \frac{Sx}{1+x}=(1+x{)}^n\frac{(1+x{)}^{n-m+1}-1}{x\cdot (1+x{)}^{n-m}}-(n-+m+1\left)\right(1+x{)}^{m-1}$
or $\frac{Sx}{1+x}=\frac{(1+x{)}^m}{x}(1+x{)}^{n-m+1}-1-(n-m+1)(1+x{)}^{m-1}$
$\therefore \frac{Sx}{1+x}=\frac{1}{x}\left[\right(1+x{)}^{n+1}-(1+x{)}^m]-(n-m+1)(1+x{)}^{m-1}$
$\therefore S=\frac{1}{x^2}\left[\right(1+x{)}^{n+2}-(1+x{)}^{m+1}]-(n-m+1)\frac{(1+x{)}^m}{x}$
$\frac{1}{x^2}(1+x{)}^{n+2}-\frac{1}{x^2}(1+x{)}^{n+2}-(n-m+1)\frac{(1+x{)}^m}{x}$
$=T_1=T_2÷T_3$There will be no terms of $x^{m}in{T}_{2}and{T}_3$as the highest degree term in both will be $x^{m-1}$The term of$x^{m}in{T}_1$ will be the term of $x^{m+2}in(1+x{)}^{n+2}$ which will be ${}^{n+2}Cm+2$