Question

If in the expansion of ${(1+x)}^n$, the coefficients of three consecutive terms are $56,70,56$, then the value of $n$ and the position of the terms of these coefficients are given by

• A. $n=8$, the terms are $4^{th},5^{th},6^{th}$
• B. $n=7$, the terms are $3^{rd},4^{th},5^{th}$
• C. $n=8$, the terms are $5^{th},6^{th},7^{th}$
• D. $n=7$, the terms are $4^{th},5^{th},6^{th}$

Answer 1

The correct option is A $n=8$, the terms are $4^{th},5^{th},6^{th}$

Let the terms be
${}^n{C}_{r-1},{}^n{C}_r,{}^n{C}_{r+1}$
Therefore from the given condition, we get
${}^n{C}_{r-1}=56$
${}^n{C}_r=70$
${}^n{C}_{r+1}=56$
Therefore
${}^n{C}_{r-1}={}^n{C}_{r+1}$
Hence $r-1=n-(r+1)$
$r-1+r+1=n$
$n=2r$
Now
$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{5}{4}$
Simplifying we get
$\frac{\left(2r\right)!}{r!r!}\times \frac{(r+1)!(r-1)!}{2r!}=\frac{5}{4}$ ...(n=2r)
$\frac{r+1}{r}=\frac{5}{4}$
Therefore $r=4$
Hence $n=8$.
The terms are ${}^8{C}_3,{}^8{C}_4,{}^8{C}_5$
$=T_4,T_5,T_6$