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

$$\begin{gathered} \mathrm{Suppose}{r}^{th},(r+1){}^{th}\mathrm{and}(r+2{)}^{th}\mathrm{terms}\mathrm{are}\mathrm{the}\mathrm{three}\mathrm{consecutive}\mathrm{terms}. \\ \mathrm{Their}\mathrm{respective}\mathrm{coefficients}\mathrm{are}{}^{n}C_{r-1},{}^{n}C_r\mathrm{and}{}^{n}C_{r+1}. \\ \mathrm{We}\mathrm{have}: \\ {}^{n}C_{r-1}={}^{n}C_{r+1}=56 \\ \Rightarrow r-1+r+1=n[\mathrm{If}{}^{n}C_r={}^{n}C_s\Rightarrow r=s\mathrm{or}r+s=n] \\ \Rightarrow 2r=n \\ \Rightarrow r=\frac{n}{2} \\ \mathrm{Now}, \\ {}^{n}C_{\frac{n}{2}}=70\mathrm{and}{}^{n}C_{\left(\frac{n}{2}-1\right)}=56 \\ \Rightarrow \frac{{}^{n}C_{\left(\frac{n}{2}-1\right)}}{{}^{n}C_{\frac{n}{2}}}=\frac{56}{70} \\ \Rightarrow \frac{{\displaystyle \frac{n}{2}}}{\left({\displaystyle \frac{n}{2}}+1\right)}=\frac{8}{10} \\ \Rightarrow 5n=4n+8 \\ \Rightarrow n=8 \\ \mathrm{So},r=\frac{n}{2}=4 \\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{terms}\mathrm{are}4\mathrm{th},5\mathrm{th}\mathrm{and}6\mathrm{th}. \end{gathered}$$