If $a_{1}, a_{2}, a_{3}, a_{4}$ be the coefficeints of four consecutive terms in the expansion of $(1 + x)^{n}$ then prove that
$\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{(a_{3} + a_{4})} = \frac{2 a_{2}}{(a_{2} + a_{3})} .$

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Solution

Let $a_{1}, a_{2}, a_{3}$ and $a_{4}$ be the coefficients of the rth, (r + 1)th, (r + 2)th
and (r + 3)th terms in the expansion of $(1 + x)^{n} .$ then,
$a_{1} =^{n} C_{r - 1}, a_{2} =_{r}^{n}, a_{3} =^{N} C_{r + 1} and a_{4} =^{n} C_{r + 2} . Now, a_{1} + a_{2} = (^{n} C_{r - 1} +^{n} C_{r}) =^{n + 1} C_{r}, a_{3} + a_{4} = (^{n} C_{r + 1} +^{n} C_{r + 2}) =^{n + 1} C_{r + 2}, and a_{2} + a_{3} = (^{n} C_{r} +^{n} C_{r + 1}) =^{n + 1} C_{r + 1} ∴ \frac{a_{1}}{(a_{1} + a_{2})} + \frac{a_{3}}{(a_{3} + a_{4})} = \frac{^{n} C_{r - 1}}{^{n + 1} C_{r}} + \frac{^{n} C_{r + 1}}{^{n + 1} C_{r + 2}} = \frac{^{n} C_{r - 1}}{(\frac{n + 1}{r} .^{n} C_{r - 1})} + \frac{n C_{r + 1}}{(\frac{n + 1}{r + 2}) .^{n} C_{r + 1}} = \frac{r}{(n + 1)} + \frac{(r + 2)}{(n + 1)} = \frac{2 (r + 1)}{(n + 1)} and, \frac{2 a_{2}}{(a_{2} + a_{3})} = \frac{2 \times^{n} C_{r}}{^{n + 1} C_{r + 1}} = \frac{2 \times^{n} C_{r}}{(\frac{n + 1}{r + 1}) .^{n} C_{r}} = \frac{2 (r + 1)}{(n + 1)} hence, \frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{(a_{3} + a_{4})} = \frac{2 a_{2}}{(a_{2} + a_{3})}$

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