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8 votes
8 votes
QUESTION 2

[16]
Given the sequence: 2; 5; 8; 11; 14...

Prove that none of the terms of this sequence are perfect squares.

PLEASE HELP!!​

User Edumelzer
by
2.4k points

2 Answers

14 votes
14 votes

Answer:

✓2=1.41421

✓5=2.23607

✓8=2.82842

✓11=3.31662

✓14=3.74166

Explanation:

Therefore the numbers are increasing by three so they cant be perfect square

User Robert Wade
by
2.6k points
16 votes
16 votes

9514 1404 393

Answer:

squares are of the form 3n or 3n+1; the sequence is of the form 3n-1, so none of the sequence will be a square

Explanation:

The given arithmetic sequence has first term 2 and common difference 3, so its explicit formula is ...

an = 2 +3(n -1) = 3n -1 . . . . for counting numbers n

__

All integers are of one of these forms: 3n-1, 3n, 3n+1, for some integer n. The squares of these are ...

(3n -1)² = 9n² -6n +1 = 3(3n² -2) +1 = 3k+1 for some k

(3n)² = 3(3n²) = 3k for some k

(3n +1)² = 9n² +6n +1 = 3(3n² +2) +1 = 3k+1 for some k

Note that none of these squares is of the form 3n -1.

Hence, the square of an integer cannot be in the given sequence.

User Coma
by
3.3k points
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