If $T_n = n(n+2)$ for a given sequence, verify that the sequence is not arithmetic.

Question

If $T_n = n(n+2)$ for a given sequence, verify that the sequence is not arithmetic.

I can't figure out how to substitute $T_{n+1}$ for $n$ correctly and work out the brackets afterwards in a good order.

Answer 1

In order to show that a sequence is not arithmetic, you just need to show that the differences between successive terms are not always the same; that is, you just need to find two terms $T_a$ and $T_b$ such that $$T_{a+1}-T_a\not=T_{b+1}-T_b.$$ That is, you're looking for a counterexample.

So let's compute some terms:

• $T_1=1(1+2)=3$.

• $T_2=2(2+2)=8$.

• $T_3=3(3+2)=15$.

Do you see how to show that the sequence is not arithmetic, now?

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Based on your comments, I suspect you are misunderstanding what "$T_n=$[stuff]" means; it means, in order to find $T_n$, plug $n$ into [stuff].

I think this is confusing because the same symbol "$n$" is getting used a lot. Think about it this way:

• Think of $T_x=x(x+2)$ as a rule: take in a number $x$, and output the number $x(x+2)$.

• So for instance, you've already calculated a few specific values of $T_x$.

• But we can also calculate examples of $T_x$ where we plug a complicated expression, involving variables,in for $x$! For example, $T_n=n(n+2)$ . . .

• And $T_{2y}=2y(2y+2)$ . . .

• And $T_{c^3+30c-2}=(c^3+30c-2)(c^3+30c-2+2)$ . . .

• and so on.

Does this help?