Question

pls solve it fast.................​

Answer 1

Solution :- (Excellent Question).

→ N = (3+1)(3²+1)(3⁴+1)(3^8 +1)__________(3^64 + 1)

Multiply and Divide by 2 in RHS , we get,

→ N = (1/2)[2 * (3+1)(3²+1)(3⁴+1)(3^8 +1)__________(3^64 + 1)]

→ N = (1/2)[(3 - 1)(3+1)(3²+1)(3⁴+1)(3^8 +1)__________(3^64 + 1)]

using (a+b)(a-b) = a² - b² now, we get,

→ N = (1/2)[(3²-1)(3²+1)(3⁴+1)(3^8 +1)__________(3^64 + 1)]

Again, using (a+b)(a-b) = a² - b² now, we get,

→ N = (1/2)[(3⁴-1)(3⁴+1)(3^8 +1)__________(3^64 + 1)]

Again, using (a+b)(a-b) = a² - b² now, we get,

→ N = (1/2)[(3^8 - 1)(3^8 +1)__________(3^64 + 1)]

So, we can see That, This Pattern Goes on ,

in Last we get,

→ N = (1/2)[(3^64 - 1)(3^64 + 1)]

using (a+b)(a-b) = a² - b² now, we get,

→ N = (1/2) [ 3^128 - 1 ]

Comparing it with Given (1/2) [ 3^a - 1 ] Now, we get,

a = 128 (Ans). (Option A).

Answer 2

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➠N = (3+1)(3²+1)(3⁴+1)(3^8+)_______(3^64+1)

➠N= (1/2)[2×(3+1)(3²+1)(3⁴+1)(3^8+1)___(3^64+1)]

➠N= (1/2)[(3-1)(3+1)(3²+1)(3⁴+1)(3^8+1)____(3^64+1)]

➠N=(1/2)[(3²-1)(3²+1)(3⁴+1)(3^8+1)____(3^64+1)

➠N= (1/2)[(3⁴-1)(3⁴+1)(3^8+1)_____(3^64+1)]

➠N= (1/2)[3^8-1)(3^8+1)_____(3^64+1)]

➠N= (1/2)[3^64-1)(3^64+1)]

➠N= (1/2) [3^128-1]

➠N= 128

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