Question

In the binomial expansion of ${(1+y)}^n$, where $n$ is a natural number, the coefficients of the $5^{th},6^{th}$ and $7^{th}$ terms are in A.P, find $n$

• A. $n=7$
• B. $n=14$
• C. $n=8$
• D. $n=16$

Answer 1

Answer: (A), (B)

$n=14$
$n=7$

The coefficient of the fifth term of the binomial expansion is ${}^n{\text{C}}_4$ , sixth term of the binomial expansion is ${}^n{\text{C}}_5$ , seventh term of the binomial expansion is ${}^n{\text{C}}_6$
Given that the $5^{th},6^{th}$ and the $7^{th}$ coefficient are in arithmetic progression,
Therefore
${}^n{\text{C}}_4$+${}^n{\text{C}}_6$ = 2${}^n{\text{C}}_5$
$\Rightarrow \frac{n!}{(n-4)!4!}+\frac{n!}{(n-6)!6!}=2\frac{n!}{(n-5)!5!}$
$\Rightarrow \frac{n(n-1)(n-2)(n-3)}{4!}+\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{\mathrm{6.5.4}!}-2$
$\Rightarrow \frac{n(n-1)(n-2)(n-3)(n-4)}{5.4!}=0$
Taking out $\frac{n(n-1)(n-2)(n-3)}{4!}$ common, we get,
$(\frac{n(n-1)(n-2)(n-3)}{4!}\left)\right(1+\frac{(n-4)(n-5)}{30}-2\frac{(n-4)}{5})=0$
Therefore,
$(1+\frac{(n-4)(n-5)}{30}-2\frac{(n-4)}{5})=0$
$\Rightarrow 30+n^2-9n+20+30-12n+48=0$
$\Rightarrow n^2-21n+98=0$
$(n-7)(n-14)=0$
So, we get $n=7$ or $14$