Leon/Samantha/Tyler -- please use the same name for your posts.
Also -- you should learn to spell CALCULUS.
If air resistance is neglected, a falling object travels 13 ft during the first second, 39 ft during the next, 65 ft during the next, and so on. These distances form the arithmetic sequence 13, 39, 65, ... .In this sequence,
a(subscript)1 = 13
d equals 26.
Find a formula for f(n), the distance fallen by the object in n seconds
f(n) = 26n-13?
F(n) = ?
4 answers
f(x) looks good
No idea what F(n) is. If you mean the sequence of partial sums, it's just the normal sum formula for A.P.'s.
No idea what F(n) is. If you mean the sequence of partial sums, it's just the normal sum formula for A.P.'s.
in general
an = a 1 + (n-1) d
so I think
f(n) = 13 + (n-1)26
that is how far it falls DURING second number n
now I bet F(n) is the total distance fallen which should approximate d = 16 t^2 if we are talking feet and seconds and g = 32 ft/s^2
that is the sum of our sequence from
n = 1 to n = t
which is
Sn = (n/2)(a1+an)
= (n/2)(a1+a1 +[n-1]d)
= (n/2) (2 a1 + n d - d )
= n a1 + n^2 d/2 - n d/2
in our case
a1 = 13 and d = 26
so
Sn = (13-26/2) n + 13 n^2
Sn = 13 n^2
well a physicist would have said (1/2)gt^2
or about 16 t^2
an = a 1 + (n-1) d
so I think
f(n) = 13 + (n-1)26
that is how far it falls DURING second number n
now I bet F(n) is the total distance fallen which should approximate d = 16 t^2 if we are talking feet and seconds and g = 32 ft/s^2
that is the sum of our sequence from
n = 1 to n = t
which is
Sn = (n/2)(a1+an)
= (n/2)(a1+a1 +[n-1]d)
= (n/2) (2 a1 + n d - d )
= n a1 + n^2 d/2 - n d/2
in our case
a1 = 13 and d = 26
so
Sn = (13-26/2) n + 13 n^2
Sn = 13 n^2
well a physicist would have said (1/2)gt^2
or about 16 t^2
By the way, on earth I think it travels 16, not 13 feet with no air resistance in the first second and d = 32, not 26
but perhaps your book is on another planet.
but perhaps your book is on another planet.
2 answers