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 #n=1#
(2) Prova that, if it holds for #n-1#, then it should be true for #n#

The first part is as easy as substituting #n=1# on #4^(2n) -1#, which gives us #4^2 - 1 = 16-1 = 15#, and #15# is indeed a multiple of #5#

The second part is a bit more tricky, and we will need to use some simple algebraic manipulation. It goes like this:
Suppose that #4^(2(n-1)) -1# is a multiple of #5#, therefore, there is a natural #k# such that #4^(2(n-1)) -1 = 5k#
Which implies
#4^(2(n-1)) = 5k+1#

#4^(2n-2) = 5k+1#

#4^(2n)*4^(-2)= 5k+1#

#4^(2n)=4^2 *(5k+1)#

#4^(2n)=16 *(5k+1)#

#4^(2n)=80k+16#

#4^(2n)-1= 80k+15#

#4^(2n)-1= 5*(16k+3)#//

And by this, we show that #4^(2n)# is also a multiple of #5# and that the second statement is also correct.
So, by the principle of induction, we have proved that the statement "#4^(2n)-1# is a multiple of #5#" is true for every natural value of #n#.

Hope it helps.

Oct 17, 2017

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 #n=1#

When #n=1# the given expression gives:

# 4^(2) - 1 = 15#

Which is divisible by #5#, So the given result is true when #n=1#.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when #n=m#, for some #m in NN, m gt 1#, in which case for this particular value of #m# we have:

# 4^(2m)-1 = 5k# ..... [B]

Where #k in NN#. Now, consider the expression [A] when we have #n=m+1#:

# 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 #5#

Induction Proof - Summary

So, we have shown that if the given result [A] is true for #n=m#, then it is also true for #n=m+1# where #m gt 1#. But we initially showed that the given result was true for #n=1# so it must also be true for #n=2, n=3, n=4, ... # and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for #n in NN#

Hence we have:

# 4^(2n)-1 # is divisible by #5 # QED