Question

If coefficient of $x^{100}$ in $1+(1+x){(1+x)}^2+.....+{(1+x)}^n(ifn\ge 100)$ is $C_{101}^{201}$ then the value of n equals

• A. $202$
• B. $100$
• C. $200$
• D. $201$

Answer 1

The correct option is C $200$

${{}^{n}C}_r+{{}^{n}C}_{(r+1)}={{}^{(n+1)}C}_{(r+1)}$
coefficient of $x^{100}$ is ${}^{100}{C}_{100}{+}^{101}{C}_{100}{+}^{102}{C}_{100}+........{+}^n{C}_{100}$.
Which is equal to ${{}^{(n+1)}C}_{101}$.
Therefore, $n+1=201$
Which implies $n=200$