Question

The coefficient of $x^7$ in $(1+2x+3x^2+\dots \dots \infty {)}^{-3}$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 0

$S=1+2x+3x^2+4x^3...\infty$........(i)
$Sx=x+2x^2+3x^3...\infty$.......(ii)
Subtracting equation (ii) from (i), we get

$S(1-x)=1+x+x^2+x^3...\infty$
$S(1-x)=(1-x{)}^{-1}$ ...(from sum of a G.P where common ration is $x$)
$S=(1-x{)}^{-2}$
Therefore, the above question can be re framed as
$(S{)}^{-3}$
$=\left(\right(1-x{)}^{-2}{)}^{-3}$
$=(1-x{)}^6$
Hence, there will be no term with $x^7$