Question

If $a_{1},a_{2},a_{3},a_{4}$ are the coefficients of any four consecutive terms in the expansion of $(1 + x)^{n}$, then $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} =$

• A. $\frac{a_{2}}{a_{2} + a_{3}}$
• B. $\frac{1}{2}\frac{a_{2}}{a_{2} + a_{3}}$
• C. $\frac{2a_{2}}{a_{2} + a_{3}}$
• D. $\frac{2a_{3}}{a_{2} + a_{3}}$

Answer 1

Answer: (C)

$\frac{2a_{2}}{a_{2} + a_{3}}$

Answer 2

Let $a_{1},a_{2},a_{3},a_{4}$ be respectively the coefficients of $(r + 1)^{th},(r + 2)^{th},(r + 3)^{th},(r + 4)^{th}$ terms in the expansion of $(1 + x)^{n}$. Then $a_{1} =^{n} ⥂ C_{r},a_{2} =^{n}C_{r + 1},a_{3} =^{n}C_{r + 2},a_{4} =^{n} ⥂ C_{r + 3}$.

Now, $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{n ⥂ C_{r}}{n ⥂ C_{r} +^{n} ⥂ C_{r + 1}} + \frac{nC_{r + 2}}{nC_{r + 2} +^{n} ⥂ C_{r + 3}}$

=$\frac{n ⥂ C_{r}}{n + 1 ⥂ C_{r + 1}} + \frac{n ⥂ C_{r + 2}}{n + 1 ⥂ C_{r + 3}}$= $\frac{n ⥂ C_{r}}{\frac{n + 1}{r + 1}\mspace{6mu}^{n}C_{r}} + \frac{nC_{r + 2}}{\frac{n + 1}{r + 3}^{n}C_{r + 2}}$

= $\frac{r + 1}{n + 1} + \frac{r + 3}{n + 1} = \frac{2(r + 2)}{n + 1}$

= $2.\frac{n ⥂ C_{r + 1}}{n + 1C_{r + 2}} = 2.\frac{n ⥂ C_{r + 1}}{nC_{r + 1} +^{n} ⥂ C_{r + 2}} = \frac{2a_{2}}{a_{2} + a_{3}}$

Let $a_{1},a_{2},a_{3},a_{4}$ be respectively the coefficients of $(r + 1)^{th},(r + 2)^{th},(r + 3)^{th},(r + 4)^{th}$ terms in the expansion of $(1 + x)^{n}$. Then $a_{1} =^{n} ⥂ C_{r},a_{2} =^{n}C_{r + 1},a_{3} =^{n}C_{r + 2},a_{4} =^{n} ⥂ C_{r + 3}$.

Now, $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{n ⥂ C_{r}}{n ⥂ C_{r} +^{n} ⥂ C_{r + 1}} + \frac{nC_{r + 2}}{nC_{r + 2} +^{n} ⥂ C_{r + 3}}$

=$\frac{n ⥂ C_{r}}{n + 1 ⥂ C_{r + 1}} + \frac{n ⥂ C_{r + 2}}{n + 1 ⥂ C_{r + 3}}$= $\frac{n ⥂ C_{r}}{\frac{n + 1}{r + 1}\mspace{6mu}^{n}C_{r}} + \frac{nC_{r + 2}}{\frac{n + 1}{r + 3}^{n}C_{r + 2}}$

=$\frac{r + 1}{n + 1} + \frac{r + 3}{n + 1} = \frac{2(r + 2)}{n + 1}$= $2.\frac{n ⥂ C_{r + 1}}{n + 1C_{r + 2}} = 2.\frac{n ⥂ C_{r + 1}}{nC_{r + 1} +^{n} ⥂ C_{r + 2}} = \frac{2a_{2}}{a_{2} + a_{3}}$