If $a_{1}, a_{2}, a_{3}$ and $a_{4}$ are the coefficients of four consecutive terms in the expansion of $(1 + x)^{n},$ 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 four consecutive terms viz rth, (r +1)th, (r + 2)th and (r + 3)the terms of the expansion $(1 + x)^{n}$ .

Then, $a_{1} = T_{r} =^{n} C_{r - 1}, a_{2} = T_{r + 1} =^{n} C_{r} .$

$a_{3} = T_{r + 2} =^{n} C_{r + 1}$ and $a_{4} = T_{r + 3} =^{n} C_{r + 2}$

Now, $a_{1} + a_{2} =^{n} C_{r - 1} +^{n} C_{r}, a_{2} + a_{3} =^{n} C_{r} +^{n} C_{r + 1} a n d a_{3} + a_{4} =^{n} C_{r + 1} +^{n} C_{r + 2}$

We have,
LHS = $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{^{n} C_{r - 1}}{^{n} C_{r - 1} +^{n} C_{r}} + \frac{^{n} C_{r + 1}}{^{n} C_{r + 1} +^{n} C_{r + 2}}$

$= \frac{^{n} C_{r - 1}}{n + 1_{C_{r}}} + \frac{^{n} C_{r + 1}}{n + 1_{C_{r + 2}}} [∵^{n} C_{r - 1} +^{n} C_{r} = {n + 1}_{C_{r}}]$

$= \frac{^{n} C_{r - 1}}{\frac{n + 1}{r} .^{n} C_{r - 1}} + \frac{^{n} C_{r + 1}}{\frac{n + 1}{r + 2} .^{n} C_{r + 1}} [∵^{n} C_{r} = \frac{n}{r} . n - 1_{C_{r - 1}}]$

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

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

$= \frac{{2.}^{n} C_{r}}{{\frac{n + 1}{r + 1}}^{n} C_{r}} = \frac{2 (r + 1)}{n + 1} [∵^{n} C_{r} +^{n} C_{r + 1} =_{C_{r + 1}}^{n + 1} . . . (i i)]$

From Eqs. (i) and (ii), we get

$\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{2 a_{2}}{a_{2} + a_{3}}$ Hence proved.

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