Question

Let $S_n{=}^n{C_0}^n{C}_1{+}^n{C_1}^n{C}_2+.....{+}^n{C_{n-1}}^n{C}_n$, where $n\in N$. If $\frac{S_{n+1}}{S_n}=\frac{15}{4}$ then the sum of all possible values of ${}^{\prime }{n}^{\prime }$ is

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is B $6$

$\begin{array}{c}{S}_n{=}^n{C_0}^n{C}_1{+}^n{C_1}^n{C}_2+.....{+}^n{C_{n-1}}^n{C}_n\\ \\ {\left(1+x\right)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+....{+}^n{C}_n{x}^n\\ \\ {\left(x+1\right)}^n{=}^n{C}_0{x}^n{+}^n{C}_1{x}^{n-1}{+}^n{C}_2{x}^{n-2}+....{+}^n{C}_n\\ \\ {\left(1+x\right)}^{2n}=\left[{}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+....{+}^n{C}_n{x}^n\right]\times \left[{}^n{C}_0{x}^n{+}^n{C}_1{x}^{n-1}{+}^n{C}_2{x}^{n-2}+....{+}^n{C}_n\right]\\ coefficientof{x}^{n-1}\\ {}^{2n}{C}_{n-1}{=}^n{C_0}^n{C}_1{+}^n{C_1}^n{C}_2+...{+}^n{C_{n-1}}^n{C}_n\\ S_n{=}^{2n}{C}_{n-1}\\ \frac{S_{n+1}}{S_n}=\frac{{}^{2n+2}{C}_n}{{}^{2n}{C}_{n-1}}=\frac{15}{4}\\ \\ \frac{\frac{\left(2n+2\right)!}{n!\left(n+2\right)!}}{\frac{2n!}{\left(n-1\right)!2n!}}=\frac{15}{4}\\ \frac{\left(2n+2\right)!}{n!\left(n+2\right)!}\times \frac{\left(n-1\right)!\times \left(n+1\right)!}{2n!}=\frac{15}{4}\\ \frac{\left(2n+2\right)\left(2n+1\right)}{n\left(n+2\right)}=\frac{15}{4}\\ 4\left(4n^2+6n+2\right)=15\left(n^2+2n\right)\\ n^2-6n+8=0\\ \left(n-4\right)\left(n-2\right)=0\\ n=2andn=4\\ sumofallvaluesofn=2+4=6\\ Hence,optionBisthecorrectanswer.\end{array}$