N-3-1 32-1 34-1 38-1 - 364-1 - If N can be simplified as 3n-1 2 then find the value of n -

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Standard VIII

Mathematics

Numbers in General Form

N=3+1 32+1 3...

Question

$N = (3 + 1) (3^{2} + 1) (3^{4} + 1) (3^{8} + 1) . (3^{64} + 1)$ . If $N$ can be simplified as $\frac{(3^{n} - 1)}{2}$ then find the value of $n$ :

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Solution

$\begin{matrix} N = (3 + 1) (3^{2} + 1) (3^{4} + 1) . . . . . . (3^{64} + 1) N = \frac{1}{(3 - 1)} [(3^{2} - 1) (3^{2} + 1) (3^{4} + 1) . . . . . . (3^{64} + 1)] = \frac{(3^{2} - 1) [(3^{2} + 1) (3^{4} + 1) . . . . . . (3^{64} + 1)]}{2} = \frac{(3^{128} - 1)}{2} H e n c e, n = 128. \end{matrix}$

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N=(3+1)(32+1)(34+1)(38+1).(364+1). If N can be simplified as (3n−1)2 then find the value of n :

N=(3+1)(32+1)(34+1)......(364+1)N=1(3−1)[(32−1)(32+1)(34+1)......(364+1)]=(32−1)[(32+1)(34+1)......(364+1)]2=(3128−1)2Hence,n=128.

$N = (3 + 1) (3^{2} + 1) (3^{4} + 1) (3^{8} + 1) . (3^{64} + 1)$ . If $N$ can be simplified as $\frac{(3^{n} - 1)}{2}$ then find the value of $n$ :

$\begin{matrix} N = (3 + 1) (3^{2} + 1) (3^{4} + 1) . . . . . . (3^{64} + 1) N = \frac{1}{(3 - 1)} [(3^{2} - 1) (3^{2} + 1) (3^{4} + 1) . . . . . . (3^{64} + 1)] = \frac{(3^{2} - 1) [(3^{2} + 1) (3^{4} + 1) . . . . . . (3^{64} + 1)]}{2} = \frac{(3^{128} - 1)}{2} H e n c e, n = 128. \end{matrix}$