If {a_1}, {a_2}, {a_3}, {a_4} are the coeffic | vedclass

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(C) Let ${a_1}, {a_2}, {a_3}, {a_4}$ be the coefficients of the $(r+1)^{th}, (r+2)^{th}, (r+3)^{th}$ and $(r+4)^{th}$ terms in the expansion of ${(1 +

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$\frac{{2{a_2}}}{{{a_2} + {a_3}}}$

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(C) Let ${a_1}, {a_2}, {a_3}, {a_4}$ be the coefficients of the $(r+1)^{th}, (r+2)^{th}, (r+3)^{th}$ and $(r+4)^{th}$ terms in the expansion of ${(1 + x)^n}$ respectively.Then ${a_1} = {^nC_r}, {a_2} = {^nC_{r+1}}, {a_3} = {^nC_{r+2}}, {a_4} = {^nC_{r+3}}$.Using the identity ${^nC_r} + {^nC_{r+1}} = {^{n+1}C_{r+1}}$,we have:$\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}} = \frac{{{^nC_r}}}{{{^{n+1}C_{r+1}}}} + \frac{{{^nC_{r+2}}}}{{{^{n+1}C_{r+3}}}}$Using the property ${^{n+1}C_{k+1}} = \frac{n+1}{k+1} {^nC_k}$,we get:$= \frac{{{^nC_r}}}{{\frac{n+1}{r+1} {^nC_r}}} + \frac{{{^nC_{r+2}}}}{{\frac{n+1}{r+3} {^nC_{r+2}}}} = \frac{r+1}{n+1} + \frac{r+3}{n+1} = \frac{2r+4}{n+1} = \frac{2(r+2)}{n+1}$.Now,consider $\frac{2{a_2}}{{a_2} + {a_3}} = \frac{2{^nC_{r+1}}}{{^nC_{r+1}} + {^nC_{r+2}}} = \frac{2{^nC_{r+1}}}{{^{n+1}C_{r+2}}} = 2 \cdot \frac{{^nC_{r+1}}}{{\frac{n+1}{r+2} {^nC_{r+1}}}} = \frac{2(r+2)}{n+1}$.Thus,$\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}} = \frac{{2{a_2}}}{{{a_2} + {a_3}}}$.

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 $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

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