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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}=

Option: 1

\frac{a_2}{a_2+a_3}


Option: 2

\frac{1}{2} \frac{a_2}{a_2+a_3}


Option: 3

\frac{2 a_2}{a_2+a_3}


Option: 4

\frac{2 a_3}{a_2+a_3}


Answers (1)

best_answer

Let a_1, a_2, a_3, a_4  be respectively the coefficients of (r+1)^{\text {th }},(r+2)^{\text {th }},(r+3)^{\text {th }},(r+4)^{\text {th }}  terms in the expansion of (1+x)^n \text {. Then } a_1={ }^n C_r, a_2={ }^n C_{r+1}, a_3={ }^n C_{r+2}, a_4={ }^n C_{r+3} \text {. }

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{{ }^n C_{r+2}}{{ }^n C_{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{\frac{n}{n+1} C_r}{\frac{n+1}{r+1} C_r}

+\frac{{ }^n C_{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 \cdot \frac{{ }^n C_{r+1}}{{ }^{n+1} C_{r+2}}=2 \cdot \frac{{ }^n C_{r+1}}{{ }^n C_{r+1}+{ }^n C_{r+2}}=\frac{2 a_2}{a_2+a_3}

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avinash.dongre

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