If some three consecutive coefficients in the binomial expansion of (x+1)^{n} in powers of x are in the ratio 2:15:70, then the average of these three coefficients is :

  • Option 1)

    964

  • Option 2)

    232

  • Option 3)

    227

  • Option 4)

    625

 

Answers (1)
V Vakul

\\n_{c_{n}}:n_{c_{r+1}}:n _{c_{r+2}}=2:15:70\\\\\:\frac{n_{c_{n}}}{n_{c_{r+1}}}=\frac{2}{15},\frac{n_{c_{r+1}}}{n_{c_{r+2}}}=\frac{15}{70},\frac{n_{c_{r}}}{n_{c_{r+2}}}=\frac{2}{70}

\frac{r+1}{n-r}=\frac{2}{15}                       \frac{r+2}{n-r-1}=\frac{3}{14}

15r+15=2n-2r\:\:\:\:\:\:\:\:\:\:\:\:14r+28=3n-3r-3

17r=2n-15\:(i)\:\:\:\:\:\:\:\:\:\:\:\:17=3n-31\:\:(ii)

(i)=(ii)

3n-31=2n-15 

n=16

putting value of  n in (i)

17r=2n-15=2\times16-15

\\17r=17\\\\\:r=1

Average =\frac{n_{c_{1}}+n_{c_{2}}+n_{c_{3}}}{3}=\frac{16_{c_{1}}+16_{c_{2}}+16_{c_{3}}}{3}

                       =\frac{16+120+560}{3}

                        =232


Option 1)

964

Option 2)

232

Option 3)

227

Option 4)

625

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