Question

lf $S_n$ denotes the sum of first '$n$' natural numbers then $S_1+S_{2}x+S_3{x}^2+\dots .\dots ..+S_n{x}^{n-1}+\dots \infty =$

• A. $(1-x{)}^{-1}$
• B. $(1-x{)}^{-2}$
• C. $(1-x{)}^{-3}$
• D. $(1-x{)}^{-4}$

Answer 1

Answer: (C)

$(1-x{)}^{-3}$

The given question can be written as.
$1+(1+2)x+(1+2+3)x^2+(1+2+3+4)x^3...\infty$
$=1+3x+6x^2+12x^3...\infty$
$(1-x{)}^{-n}=1-n(-x)+\frac{-n(-n-1)x^2}{2!}...\infty$
$=1+nx+\frac{n(n+1)x^2}{2!}...\infty$
Assuming $x=1$
Comparing the coefficients with that given in the question, we get.
$nx=3$ ...(i)
$\frac{n(n+1)x^2}{2!}=6$ ...(ii)
Substituting the value of $nx$ in equation (i), we get
$3(nx+x)=12$
$3+x=4$
$x=1$
Since $nx=3$
$n=3$
If we do not substitute the value of $x$ and only substitute the value of $n$ in the original equation, we get
$1+3x+6x^2+12x^3...\infty =(1-x{)}^{-n}$
$=(1-x{)}^{-3}$