Asked by Emily
How do you get the general term for a sequence that has increasing common differences? For example: 1,2,4,7,11
Answers
Answered by
Reiny
notice that the differences are 1,2,3,4,...
but the differences of those differences are 1,1,1,...
Since we have constants in the second set of differences our sequence can be expressed as a 2nd degree expression
(had the third set of differences been constant, it would have been a cubic expression, etc)
so let the sequence be written as
t(n) = an^2 + bn + c
if n=1 ---> a + b + c = 1
if n=2 ---> 4a + 2b + c = 2
if n=3 ---> 9a + 3b + c = 4
subtract the first two ---> 3a + b = 1
subtract the last two ---> 5a + b = 2
now subtract those
2a = 1
a = 1/2
back in 3a+b=1
3/2 + b = 1
b = -1/2
back in a+b+c=1
1/2 - 1/2 + c = 1
c=1
<b>so t(n) = (1/2)n^2 - (1/2)n + 1
= (1/2)(n^2 - n + 2)</b>
checking for t(5)
= (1/2)(25-5+2)
=(1/2)(22) = 11
but the differences of those differences are 1,1,1,...
Since we have constants in the second set of differences our sequence can be expressed as a 2nd degree expression
(had the third set of differences been constant, it would have been a cubic expression, etc)
so let the sequence be written as
t(n) = an^2 + bn + c
if n=1 ---> a + b + c = 1
if n=2 ---> 4a + 2b + c = 2
if n=3 ---> 9a + 3b + c = 4
subtract the first two ---> 3a + b = 1
subtract the last two ---> 5a + b = 2
now subtract those
2a = 1
a = 1/2
back in 3a+b=1
3/2 + b = 1
b = -1/2
back in a+b+c=1
1/2 - 1/2 + c = 1
c=1
<b>so t(n) = (1/2)n^2 - (1/2)n + 1
= (1/2)(n^2 - n + 2)</b>
checking for t(5)
= (1/2)(25-5+2)
=(1/2)(22) = 11
Answered by
Emily
why do you let n=1, 2 and 3?