Question

Answer 1

`12(5^2+1)(5^4+1)(5^8 +1)(5^16 +1) =1/2(5^n-1)(n in NN)`

`<=> 24(5^2+1)(5^4+1)(5^8 +1)(5^16 +1) =5^n-1`

`<=> (25-1)(5^2+1)(5^4+1)(5^8 +1)(5^16 +1) =5^n-1`

`<=> (5^2 -1)(5^2+1)(5^4+1)(5^8 +1)(5^16 +1) =5^n-1`

`<=> (5^4-1)(5^4+1)(5^8 +1)(5^16 +1) =5^n-1`

`<=> (5^8-1)(5^8+1)(5^16+1) =5^n-1`

`<=> (5^16-1)(5^16+1) = 5^n-1`

`<=> 5^32-1 =5^n-1`

`=> n =32(t//m)`

Thay `n=32` vào biểu thức ta đc

`2*32 -3 = 64-3 =61`

Vậy biểu thức có gt là 61 khi `n=32`

Answer 2

=>24(5^2+1)(5^4+1)(5^8+1)(5^16+1)=5^n-1

=>(5^4-1)(5^4+1)(5^8+1)(5^16+1)=5^n-1

=>5^n-1=(5^8-1)(5^8+1)(5^16+1)

=>5^n-1=5^32-1

=>n=32

=>2n-3=64-3=61