First slide

Binomial theorem for positive integral Index

Question

lf 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

Moderate

A

964
B

227
C

232
D

625

Solution

Given binomial is (x+ 1)ⁿ, whose general term, is $T_{r + 1} =^{n} C_{r} x^{r}$
According to the question, we have
$\begin{aligned} ^{n} C_{r - 1} :^{n} C_{r} :^{n} C_{r + 1} = 2 : 15 : 70 \\ now, \frac{^{n} C_{r} - 1}{^{n} C_{r}} = \frac{2}{15} \end{aligned}$
$\begin{array}{l} \Rightarrow \frac{\frac{n!}{(r - 1)! (n - r + 1)!}}{\frac{n!}{r! (n - r)!}} = \frac{2}{15} \Rightarrow \frac{r}{n - r + 1} = \frac{2}{15} \\ \Rightarrow 15 r = 2 n - 2 r + 2 \\ \Rightarrow 2 n - 17 r + 2 = 0 - - - - (i) \\ \Rightarrow similarly, \frac{^{n} C_{r}}{^{n} C_{r + 1}} = \frac{15}{70} \Rightarrow \frac{\frac{n!}{r! (n - r)!}}{\frac{n!}{(r + 1)! (n - r - 1)!}} = \frac{3}{14} \\ \Rightarrow \frac{r + 1}{n - r} = \frac{3}{14} \Rightarrow 14 r + 14 = 3 n - 3 r \\ \Rightarrow 3 n - 17 r - 14 = 0 - - - - (ii) \end{array}$
On solving Eqs. (i) and (ii), we get
$n - 16 = 0 \Rightarrow n = 16 and r = 2$
Now, the average $= \frac{^{16} C_{1} +^{16} C_{2} +^{16} C_{3}}{3}$
$= \frac{16 + 120 + 560}{3} = \frac{696}{3} = 232$

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