Question

Sum upto $n$ terms of the series $y_n=1+(1+x)+(1+x+x^2)+(1+x+x^2+x^3)+\cdots +(1+x+x^2+\cdots +x^n)$ is true for $n\in N,$ then $y_n$ is

• A. $\frac{n}{1-x}-\frac{x(1-x^n)}{(1-x{)}^2}$
• B. $\frac{n}{1-x}+\frac{x(1-x^n)}{(1-x{)}^2}$
• C. $\frac{n}{1-x}+\frac{x(1+x^n)}{(1-x{)}^2}$
• D. $\frac{-n}{1-x}+\frac{x(1-x^n)}{(1-x{)}^2}$

Answer 1

The correct option is A

$\frac{n}{1-x}-\frac{x(1-x^n)}{(1-x{)}^2}$

$n^{th}$ term of $y_n=1+x+x^2+\cdots x^n=\frac{1-x^n}{1-x}$
$y_n=\sum _{i=1}^n\frac{1-x^i}{1-x}$
$y_n=\frac{1-x}{1-x}+\frac{1-x^2}{1-x}+\frac{1-x^3}{1-x}+\cdots +\frac{1-x^n}{1-x}$
$=\frac{(1+1+1+\cdots n)-(x+x^2+\cdots \cdots +x^n)}{1-x}$
$=\frac{n}{1-x}-\frac{x(1-x^n)}{(1-x{)}^2}$

Alternate solution : if series is true for $n\in N$ is also true for $n=1,2,3,\cdots$
$\left(a\right)$ For $n=1,y_n=\frac{n}{1-x}-\frac{x(1-x^n)}{(1-x{)}^2}=\frac{1}{1-x}-\frac{x(1-x)}{(1-x{)}^2}=1$

$\left(b\right)$ For $n=1,\frac{1}{1-x}+\frac{x(1-x)}{(1-x{)}^2}\ne 1$

$\left(c\right)$ For $n=1,\frac{1}{1-x}+\frac{x(1+x)}{(1-x{)}^2}\ne 1$

$\left(d\right)$ For $n=1,\frac{-1}{1-x}+\frac{x(1-x)}{(1-x{)}^2}\ne 1$