If n is an odd positive integer, show that (n^2-1) divisible by 8.
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Thiz can b proved by taking any odd number let us take n=3
3sq-1/8=9-1/8=8/8=1
Let's take n=5
5sq-1/8=25-1/8=24/8=3
So nsq-1for any odd number iz divisible by 8
et n = 2m+1 where m=0,1,2,....
n^2=4m^2+1+4m
n^2-1=4m^2+4m
=4m(m+1)
Let K=m(m+1)
if m is odd number, odd*even will give u even number,
If m is even number even into odd will give u even number.
So K is always even no matter what is the value of m.
and even multiple of 4 will always divisble by 8
n^2=4m^2+1+4m
n^2-1=4m^2+4m
=4m(m+1)
Let K=m(m+1)
if m is odd number, odd*even will give u even number,
If m is even number even into odd will give u even number.
So K is always even no matter what is the value of m.
and even multiple of 4 will always divisble by 8
Answered by
1
take any 2 nuiimebrs
lets take 5
we get 25-1=24 which ius diuuvisible
also takje 13 we get 168 which ius divuisiuble
lets take 5
we get 25-1=24 which ius diuuvisible
also takje 13 we get 168 which ius divuisiuble
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n^2=4m^2+1+4m
n^2-1=4m^2+4m
=4m(m+1)
Let K=m(m+1)
if m is odd number, odd*even will give u even number,
If m is even number even into odd will give u even number.
So K is always even no matter what is the value of m.
and even multiple of 4 will always divisble by 8