Question

If the $2^{\text{nd}},3^{\text{rd}}$ and $4^{\text{th}}$ terms in the expansion of $(x+a{)}^n$ are $240,720$ and $1080$ respectively, then the sum of odd numbered terms in the expansion is

• A. $1664$
• B. $2376$
• C. $1562$
• D. $1486$

Answer 1

The correct option is C $1562$

$2^{\text{nd}}$ term is ${}^n{C}_1{x}^{n-1}a=240\cdots \left(1\right)$
$3^{\text{rd}}$ term is ${}^n{C}_2{x}^{n-2}{a}^2=720\cdots \left(2\right)$
$4^{\text{th}}$ term is ${}^n{C}_3{x}^{n-3}{a}^3=1080\cdots \left(3\right)$

Now, multiplying equations $\left(1\right)$ and $\left(3\right)$ and dividing by the square of equation $\left(2\right)$, we get
$\frac{{}^n{C}_1\times {}^n{C}_3}{{(}^n{C}_2{)}^2}=\frac{240\times 1080}{(720{)}^2}\Rightarrow \frac{n\times n(n-1)(n-2)(2!{)}^2}{n^2(n-1{)}^2\times 3!}=\frac{1}{2}\Rightarrow (n-2)\times 4=3\times (n-1)(\because n\ne 1)$
$\Rightarrow n=5$

Putting $n=5$ in $\left(1\right)$ and $\left(2\right)$, we get
$5x^{4}a=240$ and $10x^3{a}^2=720$
$\Rightarrow \frac{(5x^{4}a{)}^2}{10x^3{a}^2}=\frac{(240{)}^2}{720}\Rightarrow x^5=32\therefore x=2\therefore a=\frac{240}{5x^4}=\frac{48}{2^4}=3$

Now,
$(2+3{)}^5=2^5+{}^5{C}_1{2}^4\times 3+{}^5{C}_2{2}^3\times 3^2+{}^5{C}_3{2}^2\times 3^3+{}^5{C}_{4}2\times 3^4+{}^5{C}_5{3}^5=32+240+720+1080+810+243$

Hence, the sum of odd numbered terms
$=32+720+810=1562.$