You can put this solution on YOUR website! Proof is by induction:
Basis Step : If n=1
7^(2n)+16n-1 = 7^(2)+16-1 = 49+16-1 = 64 (divisible by 64)
Inductive step:
H) 7^(2n)+16n-1 is divisible by 64
T) 7^(2(n+1))+16(n+1)-1 is divisible by 64
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You can put this solution on YOUR website! Prove that 7^(2n)+16n-1 is divisible by 64
:
we prove this by using mathematical induction
:
For n=1, we have 7^2 +16 -1 = 64 which is divisible by 64
:
For n=k, we assume that 7^(2k)+16k-1 is divisible by 64
:
We must check for n=k+1, is the statement true
:
7^(2(k+1)) +16(k+1) -1 = 49 * 7^2k + 16k +15
:
we adjust this expression by adding and subtracting terms, that is
we add 49*16k, -49 but we have to subtract 49*16k and -49
:
we have the following
:
49*7^2k +16k +15 = 49*7^2k + 49*16k -49 -49*16k -49 +16k +15 =
:
49(7^2k +16k -1) - 49(16k-1) +16k +15 =
:
49(7^k +16k -1) - 48*16k +64
:
all these terms are divisible by 64, so we are done :-)
:
Here is a proof using the binomial theorem.
Every term of the binomial expansion of (8-1)2n
except the last two terms contain a factor of 8 to a power
2 or greater and thus they are all divisible by 64. So the
above becomes upon writing out the first two and last two
terms of the binomial expansion:
Simplifying and using the property of combinations C(2n,2n-1)=C(2n,1)=2n
The last two terms in the binomial expansion cancel with the
16n-1, so all the remaining terms are divisible by 64. Thus
the theorem is proved.
Edwin
