Q

# How to solve this problem- - Binomial theorem and its simple applications - JEE Main-3

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)

$625$

Views

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

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

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

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

$(i)=(ii)$

$3n-31=2n-15$

$n=16$

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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