If some three consecutive coefficients in the binomial expansion

Question:

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

  1. (1) 964

  2. (2) 232

  3. (3) 227

  4. (4) 625


Correct Option: , 2

Solution:

Given ${ }^{n} C_{r-1}:{ }^{n} C_{r}:{ }^{n} C_{r+1}=2: 15: 70$

$\Rightarrow \frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}}=\frac{2}{15}$ and $\frac{{ }^{n} C_{r}}{{ }^{n} C_{r+1}}=\frac{15}{70}$

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

$\Rightarrow 17 r=2 n+2$ and $17 r=3 n-14$

i.e., $2 n+2=3 n-14 \Rightarrow n=16 \& r=2$

$\therefore$ Average $=\frac{{ }^{16} C_{1}+{ }^{16} C_{2}+{ }^{16} C_{3}}{3}=\frac{16+120+560}{3}$

$=\frac{696}{3}=232$

Leave a comment