Binomial Theorem 1 Question 4
5.
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
(2019 Main, 9 April II)
(a) 964
(b) 227
(c) 232
(d) 625
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Answer:
Correct Answer: 5. (c)
Solution:
- Key Idea: Use general term of Binomial expansion $(x+a)^{n}$ i.e.
$T_{r+1}={ }^{n} C_{r 1} x^{n-r} a^{r}$
Given binomial is $(x+1)^{n}$, whose general term, is $T_{r+1}={ }^{n} C_{r} x^{r}$
According to the question, we have
${ }^{n} C_{r-1}:{ }^{n} C_{r}:{ }^{n} C_{r+1}=2: 15: 70$
Now, $\quad \frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}}=\frac{2}{15}$
$\Rightarrow \quad \frac{\frac{n !}{(r-1) !(n-r+1) !}} {\frac{n !}{r !(n-r) !}} =\frac{2}{15}$ $\Rightarrow \quad \frac{r}{n-r+1}=\frac{2}{15} $
$\Rightarrow \quad 15 r=2 n-2 r+2$
$\Rightarrow \quad 2 n-17 r+2=0$ $\quad$ …….(i)
Similarly, $\quad \frac{{ }^{n} C_{r}}{{ }^{n} C_{r+1}}=\frac{15}{70}$
$\Rightarrow \quad \frac {\frac{n !}{r !(n-r) !}} {\frac{n !}{(r-1) !(n-r-1) !}} =\frac{3}{14}$
$\Rightarrow \quad \frac{r+1}{n-r}=\frac{3}{14} \Rightarrow 14 r+14=3 n-3 r$
$\Rightarrow \quad 3 n-17 r-14=0$ $\quad$ …….(ii)
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 $
