Question

IF $S_n{=}^n{C}_0{.}^n{C}_1{+}^n{C}_1{.}^n{C}_2+....{+}^n{C}_{n-1}{.}^n{C}_{n}and\frac{S_{n+1}}{S_n}=\frac{15}{4}$, then n =

• A. 8
• B. 6
• C. 4
• D. 16

Answer 1

The correct option is C 4

$(1+x{)}^n{=}^n{C}_0{+}^n{C}_1^n+.....{+}^n{C}_n{n}^n$

${(1+\frac{1}{n})}^n={}^n{C}_0+{}^n{C}_1\frac{1}{n}+.....+{}^n{C}_n{\left(\frac{1}{n}\right)}^n$

$S_n$ = coefficient of n in $(1+n{)}^n{(1+\frac{1}{n})}^n$

= coefficient of n in $\frac{(1+n{)}^{2n}}{n^n}$

= coefficient of $n^{n+1}$ in $(1+n{)}^{2n}$

$={}^{2n}{C}_{n+1}$

$S_{n+1}{=}^{2n+2}{C}_{n+2}$

$\frac{s_{n+1}}{S_n}=\frac{15}{4}$

$\frac{{}^{2n}{C}_{n+1}}{{}^{2n+2}{C}_{n+2}}=\frac{15}{4}\frac{{}^{2n+2}{C}_{n+2}}{{}^{2n}{C}_{n+1}}=\frac{15}{4}$

$\frac{(2n+2)!}{(n+2)!n!}\times \frac{(n+1)!(n-1)!}{\left(2n\right)!}=\frac{15}{4}$

$\frac{(2n+2)(2n+1)}{(n+2)n}=\frac{15}{4}$

$16n^2+24n+8=15n^2+30n$

$n^2-6n+8=0$

$n-4n-2n+8=0$

$n(n-4)-2(n-4)=0$

$n=2,4$