Question

If $1+\lambda +{\lambda }^2+.....+{\lambda }^n=(1+\lambda )(1+{\lambda }^2)(1+{\lambda }^4)(1+{\lambda }^8)(1+{\lambda }^{16})$, then the value of $n$ is (where, $nϵN$)

• A. $32$
• B. $16$
• C. $31$
• D. $15$

Answer 1

The correct option is C $31$

LHS $=1+\lambda +{\lambda }^2+{\lambda }^3+....+{\lambda }^n$

$=\frac{1(1-{\lambda }^{n+1})}{(1-\lambda )}=\left(\frac{1-{\lambda }^{n+1}}{1-\lambda }\right)$ .... [Sum of first $n$ terms of G.P with c.r. $\lambda$]
RHS $=(1+\lambda )(1+{\lambda }^2)(1+{\lambda }^4)(1+{\lambda }^8)(1+{\lambda }^{16})$

$=\frac{\left[\right(1+\lambda \left)\right(1+{\lambda }^2\left)\right(1+{\lambda }^4\left)\right(1+{\lambda }^8\left)\right(1+{\lambda }^{16}\left)\right](1-\lambda )}{(1-\lambda )}$

$=\frac{\left[\right(1-{\lambda }^2\left)\right(1+{\lambda }^2\left)\right(1+{\lambda }^4\left)\right(1+{\lambda }^8\left)\right(1+{\lambda }^{16}\left)\right]}{(1-\lambda )}$

$=\frac{\left[\right(1-{\lambda }^4\left)\right(1+{\lambda }^4\left)\right(1+{\lambda }^8\left)\right(1+{\lambda }^{16}\left)\right]}{(1-\lambda )}$

$=\frac{\left[\right(1-{\lambda }^8\left)\right(1+{\lambda }^8\left)\right(1+{\lambda }^{16}\left)\right]}{(1-\lambda )}$

$=\frac{\left[\right(1-{\lambda }^{16}\left)\right(1+{\lambda }^{16}\left)\right]}{(1-\lambda )}$
$=\frac{(1-{\lambda }^{32})}{(1-\lambda )}$

$\Rightarrow \frac{1-{\lambda }^{n+1}}{1-\lambda }=\frac{(1-{\lambda }^{32})}{(1-\lambda )}$
$\Rightarrow 1-{\lambda }^{n+1}=1-{\lambda }^{32}$
$\Rightarrow n+1=32$
$\therefore n=31$

Hence, option C is correct.