Question

Solve for n

n² + 19 × n - n = 0

note - n is a natural number..​

Answer 1

Answer:

n=0 n=−18

Step-by-step explanation:

n=0 n=−18

Answer 2

Answer:

ers

ersorder by

ersorder by votes

ersorder by votesUp vote

ersorder by votesUp vote23

ersorder by votesUp vote23Down vote

ersorder by votesUp vote23Down voteAccepted

ersorder by votesUp vote23Down voteAcceptedProof by induction.

ersorder by votesUp vote23Down voteAcceptedProof by induction.Let n∈N.

ersorder by votesUp vote23Down voteAcceptedProof by induction.Let n∈N.Step 1.: Let n=1 ⇒ n<2n holds, since 1<2.

ersorder by votesUp vote23Down voteAcceptedProof 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.

ersorder by votesUp vote23Down voteAcceptedProof 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.

ersorder by votesUp vote23Down voteAcceptedProof 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.

ersorder by votesUp vote23Down voteAcceptedProof 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

ersorder by votesUp vote23Down voteAcceptedProof 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<2×2k2k<2k+1(1)

ersorder by votesUp vote23Down voteAcceptedProof 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<2×2k2k<2k+1(1)On the other hand, k>1⇒k+1<k+k=2k. Hence k+1<2k(2)

ersorder by votesUp vote23Down voteAcceptedProof 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<2×2k2k<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).

ersorder by votesUp vote23Down voteAcceptedProof 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<2×2k2k<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

ersorder by votesUp vote23Down voteAcceptedProof 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<2×2k2k<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+1k+1<2k+1

ersorder by votesUp vote23Down voteAcceptedProof 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<2×2k2k<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+1k+1<2k+1Hence, n<2n holds for all n∈N