Binomial Theorem 1 Question 24
26.
Given positive integers $r>1, n>2$ and the coefficient of $(3 r)$ th and $(r+2)$ th terms in the binomial expansion of $(1+x)^{2 n}$ are equal. Then,
(1980, 2M)
(a) $n=2 r$
(b) $n=2 r+1$
(c) $n=3 r$
(d) None of these
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Answer:
Correct Answer: 26. (a)
Solution:
- In the expansion $(1+x)^{2 n}, t_{3 r}={ }^{2 n} C_{3 r-1}(x)^{3 r-1}$
and $\quad t_{r+2}={ }^{2 n} C_{r+1}(x)^{r+1}$
Since, binomial coefficients of $t_{3 r}$ and $t_{r+2}$ are equal.
$ \begin{aligned} & \therefore \quad{ }^{2 n} C_{3 r-1}={ }^{2 n} C_{r+1} \\ & \Rightarrow \quad 3 r-1=r+1 \quad \text { or } \quad 2 n=(3 r-1)+(r+1) \\ & \Rightarrow \quad 2 r=2 \quad \text { or } \quad 2 n=4 r \\ & \Rightarrow \quad r=1 \quad \text { or } \quad n=2 r \\ & \text { But } \quad r>1 \end{aligned} $
$\therefore$ We take, $n=2 r$
