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Answer:
all work is shown and pictured
Answer:
what i don't understand why you are not answer that question
Let

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

So, the base case is ok. Now, we need to assume
and prove
.
states that

Since we're assuming
, we can substitute the sum of the first n terms with their expression:

Which terminates the proof, since we showed that

as required
D) 12 would NOT be a possible length for the third side because the sum of the 2 shorter legs should be greater than the longest leg.