Find a formula a_n
up vote
1
down vote
favorite
Find a formula an for the nth term of the arithmetic sequence whose first term is $a1=1$ such that $a_{n+1} - a_n=17$ for $n≥1$.
I am not sure on the process for solving this. Is it simply solving for $a_n$ so it would give me the result $a_n = a_{n-1} + 17$
sequences-and-series
asked Nov 17 at 0:20
Elijah
334
add a comment |
up vote
1
down vote
favorite
Find a formula an for the nth term of the arithmetic sequence whose first term is $a1=1$ such that $a_{n+1} - a_n=17$ for $n≥1$.
I am not sure on the process for solving this. Is it simply solving for $a_n$ so it would give me the result $a_n = a_{n-1} + 17$
sequences-and-series
asked Nov 17 at 0:20
Elijah
334
3
Are you sure it's geometric not arithmetic?
– KKZiomek
Nov 17 at 0:22
Oh, your right I did not notice.
– Elijah
Nov 17 at 0:26
3
You have $a_1=1$, $a_2 = 1+17$, $a_3=1+17+17=1+2cdot 17$, and so on...
– rafa11111
Nov 17 at 0:27
See if you can express $a_n$ explicitly (not in terms of any other $a_{n-1}$).
– Jam
Nov 17 at 0:33
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find a formula an for the nth term of the arithmetic sequence whose first term is $a1=1$ such that $a_{n+1} - a_n=17$ for $n≥1$.
I am not sure on the process for solving this. Is it simply solving for $a_n$ so it would give me the result $a_n = a_{n-1} + 17$
sequences-and-series
asked Nov 17 at 0:20
Elijah
334
Find a formula an for the nth term of the arithmetic sequence whose first term is $a1=1$ such that $a_{n+1} - a_n=17$ for $n≥1$.
I am not sure on the process for solving this. Is it simply solving for $a_n$ so it would give me the result $a_n = a_{n-1} + 17$
sequences-and-series
sequences-and-series
asked Nov 17 at 0:20
Elijah
334
asked Nov 17 at 0:20
Elijah
334
edited Nov 17 at 0:26
asked Nov 17 at 0:20
Elijah
334
asked Nov 17 at 0:20
Elijah
334
asked Nov 17 at 0:20
Elijah
334
334
3
Are you sure it's geometric not arithmetic?
– KKZiomek
Nov 17 at 0:22
Oh, your right I did not notice.
– Elijah
Nov 17 at 0:26
3
You have $a_1=1$, $a_2 = 1+17$, $a_3=1+17+17=1+2cdot 17$, and so on...
– rafa11111
Nov 17 at 0:27
See if you can express $a_n$ explicitly (not in terms of any other $a_{n-1}$).
– Jam
Nov 17 at 0:33
add a comment |
3
Are you sure it's geometric not arithmetic?
– KKZiomek
Nov 17 at 0:22
Oh, your right I did not notice.
– Elijah
Nov 17 at 0:26
3
You have $a_1=1$, $a_2 = 1+17$, $a_3=1+17+17=1+2cdot 17$, and so on...
– rafa11111
Nov 17 at 0:27
See if you can express $a_n$ explicitly (not in terms of any other $a_{n-1}$).
– Jam
Nov 17 at 0:33
3
3
Are you sure it's geometric not arithmetic?
– KKZiomek
Nov 17 at 0:22
Are you sure it's geometric not arithmetic?
– KKZiomek
Nov 17 at 0:22
Oh, your right I did not notice.
– Elijah
Nov 17 at 0:26
Oh, your right I did not notice.
– Elijah
Nov 17 at 0:26
3
3
You have $a_1=1$, $a_2 = 1+17$, $a_3=1+17+17=1+2cdot 17$, and so on...
– rafa11111
Nov 17 at 0:27
You have $a_1=1$, $a_2 = 1+17$, $a_3=1+17+17=1+2cdot 17$, and so on...
– rafa11111
Nov 17 at 0:27
See if you can express $a_n$ explicitly (not in terms of any other $a_{n-1}$).
– Jam
Nov 17 at 0:33
See if you can express $a_n$ explicitly (not in terms of any other $a_{n-1}$).
– Jam
Nov 17 at 0:33
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Observe that
$$
a_n=a_1+sum_{k=1}^{n-1} (a_{k+1}-a_k) quad (n>1)
$$
by telescoping sum whence
$$
a_n=1+17(n-1)quad (n>1)
$$
answered Nov 17 at 0:40
Foobaz John
19.4k41248
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Observe that
$$
a_n=a_1+sum_{k=1}^{n-1} (a_{k+1}-a_k) quad (n>1)
$$
by telescoping sum whence
$$
a_n=1+17(n-1)quad (n>1)
$$
answered Nov 17 at 0:40
Foobaz John
19.4k41248
add a comment |
up vote
0
down vote
Observe that
$$
a_n=a_1+sum_{k=1}^{n-1} (a_{k+1}-a_k) quad (n>1)
$$
by telescoping sum whence
$$
a_n=1+17(n-1)quad (n>1)
$$
answered Nov 17 at 0:40
Foobaz John
19.4k41248
add a comment |
up vote
0
down vote
up vote
0
down vote
Observe that
$$
a_n=a_1+sum_{k=1}^{n-1} (a_{k+1}-a_k) quad (n>1)
$$
by telescoping sum whence
$$
a_n=1+17(n-1)quad (n>1)
$$
answered Nov 17 at 0:40
Foobaz John
19.4k41248
Observe that
$$
a_n=a_1+sum_{k=1}^{n-1} (a_{k+1}-a_k) quad (n>1)
$$
by telescoping sum whence
$$
a_n=1+17(n-1)quad (n>1)
$$
answered Nov 17 at 0:40
Foobaz John
19.4k41248
answered Nov 17 at 0:40
Foobaz John
19.4k41248
answered Nov 17 at 0:40
Foobaz John
19.4k41248
answered Nov 17 at 0:40
Foobaz John
19.4k41248
19.4k41248
add a comment |
add a comment |
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3
Are you sure it's geometric not arithmetic?
– KKZiomek
Nov 17 at 0:22
Oh, your right I did not notice.
– Elijah
Nov 17 at 0:26
3
You have $a_1=1$, $a_2 = 1+17$, $a_3=1+17+17=1+2cdot 17$, and so on...
– rafa11111
Nov 17 at 0:27
See if you can express $a_n$ explicitly (not in terms of any other $a_{n-1}$).
– Jam
Nov 17 at 0:33