If 1+λ+λ2+…+λn =(1+λ)1+λ21+λ41+λ81+λ16 then the value of n is (where n ∊ N)

LHS =11−λn+11−λ=1−λn+11−λand RHS =(1+λ)1+λ21+λ41+λ81+λ16A=(1−λ)(1+λ)1+λ21+λ41+λ81+λ16(1−λ)=1−λ21+λ21+λ41+λ81+λ161−λ=1−λ321−λ ∴ 1−λn+11−λ =1−λ321−λ⇒ 1−λn+1 =1−λ32∴ n+1 =32⇒ n =31

If 1+λ+λ2+…+λn =(1+λ)1+λ21+λ41+λ81+λ16 then the value of n is (where n ∊ N)

LHS =11−λn+11−λ=1−λn+11−λand RHS =(1+λ)1+λ21+λ41+λ81+λ16A=(1−λ)(1+λ)1+λ21+λ41+λ81+λ16(1−λ)=1−λ21+λ21+λ41+λ81+λ161−λ=1−λ321−λ ∴ 1−λn+11−λ =1−λ321−λ⇒ 1−λn+1 =1−λ32∴ n+1 =32⇒ n =31

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If $1 + λ + λ^{2} + \dots + λ^{n}$ $= (1 + λ) (1 + λ^{2}) (1 + λ^{4}) (1 + λ^{8}) (1 + λ^{16})$ then the value of n is (where n $∊$ N)

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Detailed Solution

LHS $= \frac{1 (1 - λ^{n + 1})}{1 - λ} = \frac{1 - λ^{n + 1}}{1 - λ}$

and RHS $= (1 + λ) (1 + λ^{2}) (1 + λ^{4}) (1 + λ^{8}) (1 + λ^{16}) A$

$\begin{array}{l} = \frac{(1 - λ) (1 + λ) (1 + λ^{2}) (1 + λ^{4}) (1 + λ^{8}) (1 + λ^{16})}{(1 - λ)} \\ = \frac{(1 - λ^{2}) (1 + λ^{2}) (1 + λ^{4}) (1 + λ^{8}) (1 + λ^{16})}{1 - λ} \\ = \frac{(1 - λ^{32})}{1 - λ} \end{array}$

$\begin{array}{l} ∴ \frac{1 - λ^{n + 1}}{1 - λ} = \frac{1 - λ^{32}}{1 - λ} \\ \Rightarrow 1 - λ^{n + 1} = 1 - λ^{32} \\ ∴ n + 1 = 32 \\ \Rightarrow n = 31 \end{array}$