Question

If the largest possible coefficient of $x^{2016}$ in ${\left(1+x^{3n}+x^{504}\right)}^{10}$ (where $n\le 25,n\in N$ ) is $c$, then the value of $n+c$ is

• A. ${}^{10}{C}_4+{}^4{P}_4$
• B. ${}^{10}{C}_3+{}^6{P}_6$
• C. ${}^{11}{C}_4+{}^6{P}_6$
• D. ${}^{11}{C}_4+{}^4{P}_4$

Answer 1

Answer: (D)

${}^{10}{C}_0(1{+x^{3n})}^{10}+{}^{10}{C}_1{\left(1+x^{3n}\right)}^9\cdot x^{504}+{}^{10}{C}_2{\left(1+x^{3n}\right)}^8\cdot x^{1008}+{}^{10}{C}_3{\left(1+x^{3n}\right)}^7{x}^{1512}$
$+{}^{10}{C}_4{\left(1+x^{3n}\right)}^6{x}^{2016}+\dots \dots$
Terms containing $x^{2016}={}^{10}{C}_3\cdot {}^7{C}_r{x}^{3nr+1512}+{}^{10}{C}_4\cdot {}^6{C}_0\cdot x^{2016}$
$3nr=504\Rightarrow nr=168$
$n=\frac{168}{r},r=1,2,3,\dots \dots 6,7$
$n=24,r=7$
$\Rightarrow$ Coefficient $={}^{10}{C}_3+{}^{10}{C}_4={}^{11}{C}_4$
$\Rightarrow n+c={}^{11}{C}_4+24={}^{11}{C}_4+{}^4{P}_4$