Question

Prove that n²+2 cannot be an A.P. for every natural number.

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Answer 1

Answer:

Proof by induction.

Let n∈N.

Step 1.: Let n=1 ⇒ n<2n holds, since 1<2.

Step 2.: Assume n<2n holds where n=k and k≥1.

Step 3.: Prove n<2n holds for n=k+1 and k≥1 to complete the proof.

k<2k, using step 2.

2×k<2×2k

2k<2k+1(1)

On the other hand, k>1⇒k+1<k+k=2k. Hence k+1<2k(2)

By merging results (1) and (2).

k+1<2k<2k+1

k+1<2k+1

Hence, n<2n holds for all n∈N

Step-by-step explanation:

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Answer 2

Hint.

In an A.P there is a common difference.

Proof by contradiction: Claim the statement is true and prove it's false so that the assumption is wrong.

Solution.

Let $n^2+2$ be the sequence $\{a_{n}\}$.

Then, $a_{n+1}=(n+1)^2+2=n^2+2n+3$ and $a_{n+2}=(n+2)^2+2=n^2+4n+6$.

Let's assume to the contrary that $\{a_{n}\}$ is an A.P. Then, $d=a_{n+2}-a_{n+1}=a_{n+1}-a_{n-2}$.

Which means $d=2n+3$ by $d=a_{n+2}-a_{n+1}$, and $d=2n+1$ by $d=a_{n+1}-a_{n}$.

If there is a common difference, it would be $2n+3=2n+1$. But the equation is false for all $n$. Therefore $n^2+2$ cannot be an A.P for any $n$.