Hi, Here is another way to look at it. Consider the Arithmetic Progression(AP) in general, where we have a_(n) = a_(0) + (n-1) d , where n is the nth term of the sequence, a_(0) is the first term and d is the difference between consecutive terms.
Now, notice that in our case, the difference between consecutive terms alternates between 1 and 3 (16 - 15 = 1, 19 - 16 = 3), we could modify our formula to have d_1 (as 1) and d_2 (3) and add them alternately to the previous term. For example,
a_(1) = a_(0) + 1 (d_1) = 15 + 1 = 16 (In our case, the first term a_0 = 15 )
a_(2) = a_0 + 1 (d_1 ) + 1 (d_2 ) = 15 + 1 + 3 = 19
a_3 = a_0 + 2 (d_1 ) + 1 (d_2 ) = 15 + 2(1) + 3 = 20
a_4 = a_0 + 2(d_1 ) + 2 (d_2 ) = 15 + 2(1) + 2(3) = 23
and so on. In other words, we can write a general formula for odd terms (a_1 , a_3 ....) as follows:
a_n (n is odd) = a_0 + (n+1)/(2) (d_1 ) + (n-1)/2 (d_2 )
and for even terms (a_2 , a_4 .... ) as follows:
a_n (n is even) = + n/2 (d_1 ) + n/2 ( )
Now, we can plug in various values of n (starting from 1,2,3....) to get our required sequence.
I hope this helps.
Hello!
Consider the pairwise differences between the terms of this sequence. They are 16 - 15 = 1, 19 - 16 = 3, 20 - 19 =1, 23 - 20 = 3.
So we can suppose that the next difference will be 1, then 3, then 1 again, then 3 again and so on.
It is simple to find separate formulas for odd and even n's. They are clearly
a_n =13+2n, n is odd, a_n = 12 + 2n, n is even.
Actually, it is a legitimate formula (or at least algorithm) to find any term of the sequence. If we want to get "one formula," we can use (-1)^n, or even (1+(-1)^n)/2, which gives us the sequence 0, 1, 0, 1 and so on.
With such an addition the formula becomes
a_n = 12 + ((1+(-1)^(n+1))/2)+2n.
The next term is a_6 = 12+0+12=24.
No comments:
Post a Comment