Question

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

Answer 1

We have,

$(1+x{)}^n$

Let the three consecutive terms are $T_r,T_{r+1}$ and $T_{r+2}$

$\therefore$ Coefficients of $T_r={}^n{C}_{r-1}=56$

Coefficients of $T_{r+1}={}^n{C}_{r+1-1}={}^n{C}_r=70$ and, Coefficients of $T_{r+2}={}^n{C}_{r+2-1}={}^n{C}_{r+1}=56$

Now,

$\frac{{}^n{C}_{r+1}}{{}^n{C}_r}=\frac{56}{70}$

$\Rightarrow \frac{n-(r+1)+1}{r+1}=\frac{4}{5}$

$\Rightarrow \frac{n-r}{r+1}=\frac{4}{5}$

$\Rightarrow 5n-5r=4r+4$

$\Rightarrow 5n-9r=\mathrm{4.....}\left(i\right)$

and,

$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{70}{56}$

$\Rightarrow \frac{n-r+1}{r}=\frac{5}{4}$

$\Rightarrow 4n-4r+4=5r$

$\Rightarrow 4n-r=-4....\left(ii\right)$

Subtracting equation (ii)from (i), we get

n=4+4=8

Put n =8 in equation (i), we get

$5\times 8-9r=4$

$\Rightarrow -9r=4-40$

$\Rightarrow r=4$

$\therefore$ Three consecutive terms are 4th, 5th and 6th.