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  • If  are the coefficients of <img alt="(1+x)^n" src="https://entranceco

If a_1, a_2, a_3, a_4 are the coefficients of (1+x)^n any four consecutive terms in the expansion of , then \frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=

Option: 1

\frac{a_2}{a_2+a_3}


Option: 2

\frac{1}{2} \frac{a_2}{\left(a_2+a_3\right)}


Option: 3

\frac{2a_2}{\left(a_2+a_3\right)}


Option: 4

\frac{2a_3}{\left(a_2+a_3\right)}


Answers (1)

best_answer

Let  be respectively the coefficients of consecutive terms in the expansion of . 
Then  a_1={ }^n C_r, a_2={ }^n C_{r+1}, a_3={ }^n C_{r+2}, a_4={ }^n C_{r+3}

\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{{ }^n C_r}{{ }^n C_r+{ }^n C_{r+1}}+\frac{{ }^n C_{r+2}}{{ }^n C_{r+2}+{ }^n C_{r+3}}

\begin{aligned} & =\frac{{ }^n C_r}{{ }^{n+1} C_{r+1}}+\frac{{ }^n C_{r+2}}{{ }^{n+1} C_{r+3}}=\frac{{ }^n C_r}{\frac{n+1}{r+1} {}^n C_r}+\frac{{ }^n C_{r+2}}{\frac{n+1}{r+3}{ }^n C_{r+2}} \\ & \left.=\frac{r+1}{n+1}+\frac{r+3}{n+1}=\frac{2(r+2)}{n+1}\ \ (nC_r=\frac{n}{r}{ }^{n-1} C_{r-1}) \right. \\ & =2 \frac{{ }^n C_{r+1}}{{ }^{n+1} C_{r+2}}=2 \frac{{ }^n C_{r+1}}{{ }^n C_{r+1}+{ }^n C_{r+2}}=\frac{2 a_2}{a_2+a_3} \end{aligned}
 

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