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(B) Let the three consecutive coefficients be $^{n+2}C_{r-1}$,$^{n+2}C_{r}$,and $^{n+2}C_{r+1}$.Given the ratio $^{n+2}C_{r-1} : ^{n+2}C_{r} : ^{n+2}C

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(B) Let the three consecutive coefficients be $^{n+2}C_{r-1}$,$^{n+2}C_{r}$,and $^{n+2}C_{r+1}$.Given the ratio $^{n+2}C_{r-1} : ^{n+2}C_{r} : ^{n+2}C_{r+1} = 1 : 3 : 5$.From $\frac{^{n+2}C_{r-1}}{^{n+2}C_{r}} = \frac{1}{3}$,we get $\frac{r}{n-r+3} = \frac{1}{3} \implies 3r = n-r+3 \implies n = 4r-3$ $(i)$.From $\frac{^{n+2}C_{r}}{^{n+2}C_{r+1}} = \frac{3}{5}$,we get $\frac{r+1}{n-r+2} = \frac{3}{5} \implies 5r+5 = 3n-3r+6 \implies 3n = 8r-1$ $(ii)$.Substituting $n = 4r-3$ into $(ii)$: $3(4r-3) = 8r-1 \implies 12r-9 = 8r-1 \implies 4r = 8 \implies r = 2$.Then $n = 4(2)-3 = 5$.The coefficients are $^{7}C_{1}, ^{7}C_{2}, ^{7}C_{3}$,which are $7, 21, 35$.The sum is $7 + 21 + 35 = 63$.

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+2}$,which are in the ratio $1:3:5$,is equal to

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