Prove that #4^(2n)-1 # is divisible by #5 AA n in NN#?
2 Answers
In the explanation...
Explanation:
To prove this statement by induction, we just have to follow these two steps:
(1) Prove that it holds for
(2) Prova that, if it holds for
The first part is as easy as substituting
The second part is a bit more tricky, and we will need to use some simple algebraic manipulation. It goes like this:
Suppose that
Which implies
And by this, we show that
So, by the principle of induction, we have proved that the statement "
Hope it helps.
Induction Proof - Hypothesis
We seek to prove that the the expression:
# 4^(2n)-1 # is divisible by#5 AA n in NN# ..... [A]
So let us test this assertion using Mathematical Induction:
Induction Proof - Base case:
We will show that the given result, [A], holds for
When
# 4^(2) - 1 = 15#
Which is divisible by
Induction Proof - General Case
Now, Let us assume that the given result [A] is true when
# 4^(2m)-1 = 5k# ..... [B]
Where
# 4^(2(m+1))-1 -= 4^(2m+2)-1 #
# " " = 4^(2m)4^2-1 #
# " " = (5k+1)4^2-1 # using [B]
# " " = 16(5k+1) - 1 #
# " " = 16.5k+16 - 1 #
# " " = 16.5k+15 #
# " " = 5(16k+3) #
Which is also divisible by
Induction Proof - Summary
So, we have shown that if the given result [A] is true for
Induction Proof - Conclusion
Then, by the process of mathematical induction the given result [A] is true for
Hence we have:
# 4^(2n)-1 # is divisible by#5 # QED