Question

prove that (5^n )-4n +1 is divisble by 16 for n ={0, 1, 2, 3,...n, n+1}

Answer 1

plz chec k the question.. it may be (5^n )-4n -1
if so express it as (4+1)^n -4n -1 and use binomial theorems then 4^2 will come as factor

Answer 2

correction n={1, 2, .... n, n+1}

Answer 3

still quest is not correct it may be (5^n )-4n -1

Answer 4

I think it is 5^n-4n-1 and n={2,3,4,5...}

Answer 5

if n starts from 1 that will also do

Answer 6

ok, 0/16=0

Answer 7

yup

Answer 8

we use method inductive

Answer 9

induction can also be used

Answer 10

n=0 it is divisble by 16
suppose n=k it true
define 5^k-4k-1 divisble by 16
we prove n=k+1 true
5^(k+1)-4(k+1)-1=5*5^k-4k-4-1=5*(5^k-4k-1)+16k
we have 5*(5^k-4k-1) divisble by 16 and 16k divisble by 16
thus
5*(5^k-4k-1)+16k divisble by 16

Answer 11

yes the ? is (5^n) -4n-1 ,
i am confused using binomial thereom mainly, because i have never used it.

Answer 12

use P.M.I. instead

Answer 13

Just to make what chanh wrote a bit easier to read, we have \[\large 5^{n+1}-4(n+1)-1=5^{n+1}-4n-5\]\[=5^{n+1}-20n-5+16n\]\[=5(5^n-4n-1)+16n\]Both sides are divisible by 16, so we're done.

Answer 14

I use induction

Answer 15

so you are allowed to only multiply -4n by 5 and to add 16n,

Answer 16

KingGeorge explained
we have -20n+16n=-4n

Answer 17

oh, i see