Question

Find the two successive terms in the expansion of $(2+3x{)}^8$ whose coefficients are in the ratio 1:3.

• A. 2nd and 3rd terms
• B. 3rd and 4th terms
• C. 4th and 5th terms
• D. 5th and 6th terms

Answer 1

The correct option is B 3rd and 4th terms

Given,

$(2+3x{)}^8$

general term,

$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

$T_{r+1}={}^8{C}_r\left(2{)}^{8-r}\right(3x{)}^r$

let the terms be $T_{r+1}$ and $T_{r+2}$

$T_{r+1}={}^8{C}_r\left(2{)}^{8-r}\right(3x{)}^r$

$T_{r+1+1}={}^8{C}_{r+1}\left(2{)}^{8-r-1}\right(3x{)}^{r+1}$

$T_{r+1}={}^8{C}_r(2{)}^{8-r}{3}^r{x}^r$

$T_{r+2}={}^8{C}_{r+1}\left(2{)}^{7-r}\right(3{)}^{r+1}(x{)}^{r+1}$

$\frac{{}^8{C}_r(2{)}^{8-r}{3}^r}{{}^8{C}_{r+1}\left(2{)}^{7-r}\right(3{)}^{r+1}}=\frac{1}{3}$

$\frac{8!(r+1)!(8-r-1)!(2{)}^1}{r!(8-r)!\times 8!\left(3\right)}=\frac{1}{3}$

$\frac{(r+1)\left(r\right)!(7-r)!2}{r!(8-r)(7-r)!\times 3}=\frac{1}{3}$

$2(r+1)=8-r$

$2r+2=8-r$

$3r=6$

$r=2$

$r+1=3$ and $r+2=4$

So these are $3^{rd}$ and $4^{th}$ terms.