The coefficients of three consecutive terms o | vedclass

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(B) Let the three consecutive terms be $T_r, T_{r+1}, T_{r+2}$. Their coefficients are ${}^{n+5}C_{r-1}, {}^{n+5}C_r, {}^{n+5}C_{r+1}$.Given the ratio

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$6$

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(B) Let the three consecutive terms be $T_r, T_{r+1}, T_{r+2}$. Their coefficients are ${}^{n+5}C_{r-1}, {}^{n+5}C_r, {}^{n+5}C_{r+1}$.Given the ratio ${}^{n+5}C_{r-1} : {}^{n+5}C_r : {}^{n+5}C_{r+1} = 5 : 10 : 14$.From $\frac{{}^{n+5}C_r}{{}^{n+5}C_{r-1}} = \frac{10}{5} = 2$,we get $\frac{(n+5)-r+1}{r} = 2 \Rightarrow n+6 = 3r$.From $\frac{{}^{n+5}C_{r+1}}{{}^{n+5}C_r} = \frac{14}{10} = \frac{7}{5}$,we get $\frac{(n+5)-r}{r+1} = \frac{7}{5}$ $\Rightarrow 5n + 25 - 5r = 7r + 7$ $\Rightarrow 5n + 18 = 12r$.Substituting $r = \frac{n+6}{3}$ into the second equation: $5n + 18 = 12(\frac{n+6}{3}) = 4(n+6) = 4n + 24$.Thus,$n = 24 - 18 = 6$.

The coefficients of three consecutive terms of $(1+x)^{n+5}$ are in the ratio $5: 10: 14$. Then $n=$

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