Can You Find the Sequence?

Geometry Level 2

2 , 12 , 14 , 24 , 26 , 36 , 38 , 48 , 50 , . . . a n 2,\: 12, \: 14, \: 24, \: 26, \: 36, \: 38, \: 48, \: 50,... \: a_{n}

Which of the following is a possible formula of a n a_{n} for the above sequence?

Hint: don't just guess and check!

2 n 2 + abs [ 10 sin ( ( n + 1 ) π 2 ) ] 2n-2+\textrm{abs}\left [ 10\sin \left (\frac{(n+1)\pi}{2} \right ) \right ] 6 n + 2 cos ( π n ) 2 6n+2\cos(\pi n)-2 4 n + 2 cos ( π n ) 4n+2\cos(\pi n) 10 n 2 10n-2

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1 solution

Marta Reece
Apr 11, 2017

The sequence can be obtained by adding + 2 , + 10 , + 2 , + 10 +2, +10, +2, +10 , etc. to each term in turn.

So we need to increase a n a_n by 12 12 for n n increasing by 2 2 . 6 n 6n will accomplish that.

Next we need to deviate from + 6 , + 6 +6, +6 , etc. down and then up, and down and up. If we deviate by 2 2 down and then 2 2 up, that will give us 4 4 change in total, that is instead of + 6 , + 6 +6,+6 , we will have + 6 4 , + 6 + 4 +6-4, +6+4 , or + 2 , + 10 +2, +10 as we want.

To get 2 2 down 2 2 up pattern, we can use 2 × c o s ( π n ) 2\times cos(\pi n) .

Starting with n = 1 n=1 our sequence at this point is: 4 , 14 , 16 4, 14, 16 etc. So we need to subtract 2 2 to get the numbers in the original sequence.

The answer is 6 n + 2 c o s ( π n ) 2 6n+2cos(\pi n)-2

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