Prove the following statements using mathematical induction.
We shall prove this using induction.
In the basis step, , we see that
and
and so the basis step holds.
In the inductive step, we will assume that for some positive integer and show that . By the inductive hypothesis,
With some algebraic manipulation this becomes
or
and so the inductive step holds.
Since the inductive step and the basis step hold, it is true that for every positive integer .
We shall prove this using induction.
In the basis step, , we see that
and
and so the basis step holds.
In the inductive step, we will assume that for some positive integer and show that . By the inductive hypothesis,
Factoring this yields
and so the inductive step holds.
Since the inductive step and the basis step hold, it is true that for every positive integer . ∎
We shall prove this using induction.
In the basis step, , we see that
and so the basis step holds.
In the inductive step, we will assume for some positive integer and show that . Applying the inductive hypothesis,
Note that for ,
so
By substitution,
and hence the inductive step holds.
Since the inductive step and the basis step hold, for every positive n that is greater than 4. ∎
There are many different ways to show
for - it may be useful to try induction here.
We shall prove this using induction.
In the basis step, ,
so the basis step holds.
In the inductive step, we assume is divisible by 5 for some positive integer and we will show is divisible by 5. Expanding the left-hand side yields,
or, combining like terms except for ,
Since this is the sum of two integers which are divisble by five, is divisible by 5. Hence, the inductive step holds.
Since the inductive step and the basis step hold, is divisible by 5 for every positive integer n. ∎
We shall prove this using induction.
In the basis step, ,
and
so the basis step holds.
In the inductive step, we assume for some positive integer and we will show . Applying the inductive hypothesis to the left-hand side yields
or
which gives
so
and hence the inductive step holds.
Since the inductive step and the basis step hold, for every positive integer n. ∎
The formula is . The proof is left to the reader.
The sum is . The proof is left to the reader.