Question:
If some three consecutive in the binomial expansion of $(x + 1)^n$ is powers of $x$ are in the ratio $2 : 15 : 70$, then the average of these three coefficient is :-
If some three consecutive in the binomial expansion of $(x + 1)^n$ is powers of $x$ are in the ratio $2 : 15 : 70$, then the average of these three coefficient is :-
Updated On: Apr 2, 2026
- 964
- 625
- 227
- 232
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The Correct Option is D
Solution and Explanation
$\frac{^{n}C_{r-1}}{^{n}C_{r}} = \frac{2}{15} $
$ \frac{\frac{n!}{\left(r-1\right)!\left(n-r+1\right)!}}{\frac{n!}{r!\left(n-r\right)!}} = \frac{2}{15} $
$ \frac{r}{n-r+1} = \frac{2}{15} $
$ 15r =2n-2r+2 $
$ 17r = 2n+2 $
$ \frac{^{n}C_{r}}{^{n}C_{r+1}} = \frac{15}{70} $
$ \frac{\frac{n!}{r!\left(n-r\right)!}}{\frac{n!}{\left(r+1\right)!\left(n-r-1\right)!}} = \frac{3}{14}$
$ \frac{r+1}{n-r} = \frac{3}{14} $
$ 14r +14 =3n -3r $
$ 3n - 17r =14$
$ \frac{2n-17r =- 2 }{n=16} $
$ 17r=34, r = 2 $
${^{16}C_{1}} , {^{16}C_{2}} , {^{16}C_{3}} $
$ \frac{^{16}C_{1} + ^{16}C_{2} + ^{16}C_{3}}{3 } = \frac{16+120+560}{3} $
$\frac{680+16}{3} = \frac{696}{3} = 232 $
$ \frac{\frac{n!}{\left(r-1\right)!\left(n-r+1\right)!}}{\frac{n!}{r!\left(n-r\right)!}} = \frac{2}{15} $
$ \frac{r}{n-r+1} = \frac{2}{15} $
$ 15r =2n-2r+2 $
$ 17r = 2n+2 $
$ \frac{^{n}C_{r}}{^{n}C_{r+1}} = \frac{15}{70} $
$ \frac{\frac{n!}{r!\left(n-r\right)!}}{\frac{n!}{\left(r+1\right)!\left(n-r-1\right)!}} = \frac{3}{14}$
$ \frac{r+1}{n-r} = \frac{3}{14} $
$ 14r +14 =3n -3r $
$ 3n - 17r =14$
$ \frac{2n-17r =- 2 }{n=16} $
$ 17r=34, r = 2 $
${^{16}C_{1}} , {^{16}C_{2}} , {^{16}C_{3}} $
$ \frac{^{16}C_{1} + ^{16}C_{2} + ^{16}C_{3}}{3 } = \frac{16+120+560}{3} $
$\frac{680+16}{3} = \frac{696}{3} = 232 $
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