Question

The coefficient of $x^4$ in the binomial expansion of $(1+x{)}^4+(1+x{)}^5+\cdots +(1+x{)}^{11}+(1+bx{)}^{12}$ is $(n+1)\cdot {}^{13}{C}_5$ for some $n\in N$. Then the smallest natural number possible for $b$ is

Answer 1

$(1+x{)}^4+(1+x{)}^5+\cdots +(1+x{)}^{11}+(1+bx{)}^{12}$
$=\frac{(1+x{)}^4[(1+x{)}^8-1]}{(1+x)-1}+(1+bx{)}^{12}$
$=\frac{(1+x{)}^{12}}{x}-\frac{(1+x{)}^4}{x}+(1+bx{)}^{12}$

Coefficient of $x^4={}^{12}{C}_5-0+{}^{12}{C}_4{b}^4$
So, ${}^{12}{C}_5+{}^{12}{C}_4{b}^4=(n+1)\cdot {}^{13}{C}_5$
$\Rightarrow b^4-1=\frac{13}{5}n$
$\Rightarrow n=\frac{5(b^4-1)}{13}$
$\Rightarrow n=\frac{5(b-1)(b+1)(b^2+1)}{13}$
Since $n\in N$, the least possible value of $b$ is $12$