Need help :P (1 Viewer)

[hon1hon2hon3]

show (3^4n ) minus 1 is divisible by 80 for all positive intergral value of n

Need helpp mental blockage >_>

any one give me a pump ?

oh sheeeze lol, right after this post , i found out how to do it lol . . .

For the last bit, K= K +1 , u will get (3^4(k+1)) -1 = (3^4 times 3^4k) -1 = (81 times 3^4k minus 1) = ( 80 times 3^4k plus 3^4k minus 1 )

and they are both divisible by 80 , thus ture . . .

wtf = = i am soo random , i answered my own question = = just ignore me = =" sorry for wasting a space ="=

 

[lyounamu]

show (3^4n ) minus 1 is divisible by 80 for all positive intergral value of n

Need helpp mental blockage >_>

any one give me a pump ?

oh sheeeze lol, right after this post , i found out how to do it lol . . .

For the last bit, K= K +1 , u will get (3^4(k+1)) -1 = (3^4 times 3^4k) -1 = (81 times 3^4k minus 1) = ( 80 times 3^4k plus 3^4k minus 1 )

and they are both divisible by 80 , thus ture . . .

wtf = = i am soo random , i answered my own question = = just ignore me = =" sorry for wasting a space ="=

When n =1, 3^4n - 1 = 80 which is dividible by 80.

Now assume it is true for n=k
i.e. 3^4k - 1 = 80m where m is any positive integer.

Hence prove it is also true for n=k+1
i.e. 3^(4k+4) - 1 = 80T where T is another any positive integer.
81 . 3^4k - 1 = 81 . 3^4k - 81 + 80
= 81(3^4k - 1) + 80
= 81 . 80m + 80
= 80 (81m+1) where 81m + 1 is any another integer = T.

 

[hon1hon2hon3]

thx for the reply , but i have already finished t he questiosn thx any way