The coefficients of the (r-1)^{th},r^{th},and | vedclass

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(A) The $(k+1)^{th}$ term in the binomial expansion of $(x+1)^{n}$ is given by $T_{k+1} = {^nC_k} x^{n-k}$.The $(r-1)^{th}$ term is $T_{r-1} = {^nC_{r

Vedclass · Accepted Answer

$n=7, r=3$

Vedclass · Answer

(A) The $(k+1)^{th}$ term in the binomial expansion of $(x+1)^{n}$ is given by $T_{k+1} = {^nC_k} x^{n-k}$.The $(r-1)^{th}$ term is $T_{r-1} = {^nC_{r-2}} x^{n-r+2}$,so its coefficient is ${^nC_{r-2}}$.The $r^{th}$ term is $T_r = {^nC_{r-1}} x^{n-r+1}$,so its coefficient is ${^nC_{r-1}}$.The $(r+1)^{th}$ term is $T_{r+1} = {^nC_r} x^{n-r}$,so its coefficient is ${^nC_r}$.Given the ratio of coefficients is $1:3:5$,we have:$\frac{^nC_{r-2}}{^nC_{r-1}} = \frac{1}{3}$ and $\frac{^nC_{r-1}}{^nC_r} = \frac{3}{5}$.From $\frac{^nC_{r-2}}{^nC_{r-1}} = \frac{r-1}{n-r+2} = \frac{1}{3}$,we get $3r-3 = n-r+2$,which simplifies to $n-4r+5=0$ (Equation $1$).From $\frac{^nC_{r-1}}{^nC_r} = \frac{r}{n-r+1} = \frac{3}{5}$,we get $5r = 3n-3r+3$,which simplifies to $3n-8r+3=0$ (Equation $2$).Multiplying Equation $1$ by $2$,we get $2n-8r+10=0$. Subtracting this from Equation $2$ $(3n-8r+3=0)$,we get $n-7=0$,so $n=7$.Substituting $n=7$ into Equation $1$: $7-4r+5=0$ $\Rightarrow 4r=12$ $\Rightarrow r=3$.Thus,$n=7$ and $r=3$.

The coefficients of the $(r-1)^{th}$,$r^{th}$,and $(r+1)^{th}$ terms in the expansion of $(x+1)^{n}$ are in the ratio $1:3:5$. Find $n$ and $r$.

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