Question

If $S$ be the sum of the coefficients in the expansion of $(ax+by+-cz{)}^n$ where $a,b,c$ are lengths of the sides of a triangle, then ${lim}_{n\to \infty }\frac{S}{(S^{1/n}+1{)}^n}$ is

• A. $1$
• B. $0$
• C. $e^{\left(\frac{ab}{c}\right)}$
• D. $e^{\left(\frac{a+b+c}{a+b+c-1}\right)}$

Answer 1

Answer: (B)

$0$

$S=(a+b-c{)}^n$
$\underset{n\to \infty }{lim}\frac{S}{{(S^{1/n}+1)}^n}$ = $\underset{n\to \infty }{lim}{\left(\frac{a+b-c}{a+b-c+1}\right)}^n$
$=\underset{n\to \infty }{lim}{\left(\frac{k}{k+1}\right)}^n=L$ (say)
Where $k=a+b-c$
Now In $\Delta ABCa+b>c$ $\Rightarrow 0<\frac{k}{k+1}<1\therefore L=0$