Mixing
Sum of series: 9 + 16 + 29 + 54 + 103 + …
$begingroup$
The series is neither AP nor GP, so I tried doing some manipulations to get the expression for the general term.
$S = 9 + 16 + 29 + 54 + 103 + ... T_n$
$S = 0 + 9 + 16 + 29 + 54 + ... T_{n-1} + T_n$
Subtracting the two:
$0 = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1}) - T_n$
This gives:
$T_n = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1})$
The expression obtained for $T_n$ is neither in AP nor in GP.
I repeated the above process with $T_n$ (which gave me a GP this time) and got this expression finally:
$T_n - T_{n-1} = 7 + 6(2^{n-2} - 1) = 6(2^{n-2}) + 1$
How do I proceed from here to get an expression for the general term as well as the sum of the series?
sequences-and-series algebra-precalculus
asked Jan 26 at 15:56
user638500user638500
403
$endgroup$
add a comment |
$begingroup$
The series is neither AP nor GP, so I tried doing some manipulations to get the expression for the general term.
$S = 9 + 16 + 29 + 54 + 103 + ... T_n$
$S = 0 + 9 + 16 + 29 + 54 + ... T_{n-1} + T_n$
Subtracting the two:
$0 = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1}) - T_n$
This gives:
$T_n = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1})$
The expression obtained for $T_n$ is neither in AP nor in GP.
I repeated the above process with $T_n$ (which gave me a GP this time) and got this expression finally:
$T_n - T_{n-1} = 7 + 6(2^{n-2} - 1) = 6(2^{n-2}) + 1$
How do I proceed from here to get an expression for the general term as well as the sum of the series?
sequences-and-series algebra-precalculus
asked Jan 26 at 15:56
user638500user638500
403
$endgroup$
add a comment |
$begingroup$
The series is neither AP nor GP, so I tried doing some manipulations to get the expression for the general term.
$S = 9 + 16 + 29 + 54 + 103 + ... T_n$
$S = 0 + 9 + 16 + 29 + 54 + ... T_{n-1} + T_n$
Subtracting the two:
$0 = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1}) - T_n$
This gives:
$T_n = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1})$
The expression obtained for $T_n$ is neither in AP nor in GP.
I repeated the above process with $T_n$ (which gave me a GP this time) and got this expression finally:
$T_n - T_{n-1} = 7 + 6(2^{n-2} - 1) = 6(2^{n-2}) + 1$
How do I proceed from here to get an expression for the general term as well as the sum of the series?
sequences-and-series algebra-precalculus
asked Jan 26 at 15:56
user638500user638500
403
$endgroup$
The series is neither AP nor GP, so I tried doing some manipulations to get the expression for the general term.
$S = 9 + 16 + 29 + 54 + 103 + ... T_n$
$S = 0 + 9 + 16 + 29 + 54 + ... T_{n-1} + T_n$
Subtracting the two:
$0 = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1}) - T_n$
This gives:
$T_n = 9 + 7 + 13 + 25 + 49 + ... (T_n - T_{n-1})$
The expression obtained for $T_n$ is neither in AP nor in GP.
I repeated the above process with $T_n$ (which gave me a GP this time) and got this expression finally:
$T_n - T_{n-1} = 7 + 6(2^{n-2} - 1) = 6(2^{n-2}) + 1$
How do I proceed from here to get an expression for the general term as well as the sum of the series?
sequences-and-series algebra-precalculus
sequences-and-series algebra-precalculus
asked Jan 26 at 15:56
user638500user638500
403
asked Jan 26 at 15:56
user638500user638500
403
asked Jan 26 at 15:56
user638500user638500
403
asked Jan 26 at 15:56
user638500user638500
403
asked Jan 26 at 15:56
user638500user638500
403
403
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
From $T_n - T_{n-1} = 6(2^{n-2}) + 1$ you have
$$T_n = T_0 + sum_{k=1}^n (6(2^{n-2}) + 1) \
= T_0 + n + 3 (2^n -1)$$
The general approach is:
$$T_n = a+ n + b 2^n$$
so
$$9 = T_1 = a + 1 + 2 b\
16 = T_2 = a + 2 + 4b
$$
which gives indeed
$b = 3$ and $a = 2$, hence
$$T_n = 2+ n + 3 cdot 2^n$$
The sum is then
$$
S_n = sum_{k=1}^n T_k = 2 n + frac12 n (n+1) + 3 (2^{n+1} -1)
$$
answered Jan 26 at 16:15
AndreasAndreas
8,3511137
$endgroup$
$begingroup$
Can you please explain what $T_0$ is and what happened to $T_{n-1}$?
$endgroup$
– user638500
Jan 26 at 17:14
$begingroup$
$T_0$ is some start value which we don't know a priori, since we only identified differences. $T_{n-1}$ disappears by telescoping: $T_n = T_n - T_{n-1} + T_{n-1} - T_{n-2} + ... + T_{2} - T_{1} + T_0$. You regroup that to $T_n = (T_n - T_{n-1}) + (T_{n-1} - T_{n-2}) + ... + (T_{2} - T_{1}) + T_0$ which gives you the fomula for $T_n$.
$endgroup$
– Andreas
Jan 27 at 12:29
add a comment |
$begingroup$
Now if you define $R_n=T_n+n$ you can get the recurrence $$R_n-R_{n-1}=6cdot 2^{n-2}\R_1=T_1-1=8$$
and you get an arithmetico-geometric series for $R$
answered Jan 26 at 16:06
Ross MillikanRoss Millikan
300k24200374
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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oldest
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active
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$begingroup$
From $T_n - T_{n-1} = 6(2^{n-2}) + 1$ you have
$$T_n = T_0 + sum_{k=1}^n (6(2^{n-2}) + 1) \
= T_0 + n + 3 (2^n -1)$$
The general approach is:
$$T_n = a+ n + b 2^n$$
so
$$9 = T_1 = a + 1 + 2 b\
16 = T_2 = a + 2 + 4b
$$
which gives indeed
$b = 3$ and $a = 2$, hence
$$T_n = 2+ n + 3 cdot 2^n$$
The sum is then
$$
S_n = sum_{k=1}^n T_k = 2 n + frac12 n (n+1) + 3 (2^{n+1} -1)
$$
answered Jan 26 at 16:15
AndreasAndreas
8,3511137
$endgroup$
$begingroup$
Can you please explain what $T_0$ is and what happened to $T_{n-1}$?
$endgroup$
– user638500
Jan 26 at 17:14
$begingroup$
$T_0$ is some start value which we don't know a priori, since we only identified differences. $T_{n-1}$ disappears by telescoping: $T_n = T_n - T_{n-1} + T_{n-1} - T_{n-2} + ... + T_{2} - T_{1} + T_0$. You regroup that to $T_n = (T_n - T_{n-1}) + (T_{n-1} - T_{n-2}) + ... + (T_{2} - T_{1}) + T_0$ which gives you the fomula for $T_n$.
$endgroup$
– Andreas
Jan 27 at 12:29
add a comment |
$begingroup$
From $T_n - T_{n-1} = 6(2^{n-2}) + 1$ you have
$$T_n = T_0 + sum_{k=1}^n (6(2^{n-2}) + 1) \
= T_0 + n + 3 (2^n -1)$$
The general approach is:
$$T_n = a+ n + b 2^n$$
so
$$9 = T_1 = a + 1 + 2 b\
16 = T_2 = a + 2 + 4b
$$
which gives indeed
$b = 3$ and $a = 2$, hence
$$T_n = 2+ n + 3 cdot 2^n$$
The sum is then
$$
S_n = sum_{k=1}^n T_k = 2 n + frac12 n (n+1) + 3 (2^{n+1} -1)
$$
answered Jan 26 at 16:15
AndreasAndreas
8,3511137
$endgroup$
$begingroup$
Can you please explain what $T_0$ is and what happened to $T_{n-1}$?
$endgroup$
– user638500
Jan 26 at 17:14
$begingroup$
$T_0$ is some start value which we don't know a priori, since we only identified differences. $T_{n-1}$ disappears by telescoping: $T_n = T_n - T_{n-1} + T_{n-1} - T_{n-2} + ... + T_{2} - T_{1} + T_0$. You regroup that to $T_n = (T_n - T_{n-1}) + (T_{n-1} - T_{n-2}) + ... + (T_{2} - T_{1}) + T_0$ which gives you the fomula for $T_n$.
$endgroup$
– Andreas
Jan 27 at 12:29
add a comment |
$begingroup$
From $T_n - T_{n-1} = 6(2^{n-2}) + 1$ you have
$$T_n = T_0 + sum_{k=1}^n (6(2^{n-2}) + 1) \
= T_0 + n + 3 (2^n -1)$$
The general approach is:
$$T_n = a+ n + b 2^n$$
so
$$9 = T_1 = a + 1 + 2 b\
16 = T_2 = a + 2 + 4b
$$
which gives indeed
$b = 3$ and $a = 2$, hence
$$T_n = 2+ n + 3 cdot 2^n$$
The sum is then
$$
S_n = sum_{k=1}^n T_k = 2 n + frac12 n (n+1) + 3 (2^{n+1} -1)
$$
answered Jan 26 at 16:15
AndreasAndreas
8,3511137
$endgroup$
From $T_n - T_{n-1} = 6(2^{n-2}) + 1$ you have
$$T_n = T_0 + sum_{k=1}^n (6(2^{n-2}) + 1) \
= T_0 + n + 3 (2^n -1)$$
The general approach is:
$$T_n = a+ n + b 2^n$$
so
$$9 = T_1 = a + 1 + 2 b\
16 = T_2 = a + 2 + 4b
$$
which gives indeed
$b = 3$ and $a = 2$, hence
$$T_n = 2+ n + 3 cdot 2^n$$
The sum is then
$$
S_n = sum_{k=1}^n T_k = 2 n + frac12 n (n+1) + 3 (2^{n+1} -1)
$$
answered Jan 26 at 16:15
AndreasAndreas
8,3511137
answered Jan 26 at 16:15
AndreasAndreas
8,3511137
answered Jan 26 at 16:15
AndreasAndreas
8,3511137
answered Jan 26 at 16:15
AndreasAndreas
8,3511137
8,3511137
$begingroup$
Can you please explain what $T_0$ is and what happened to $T_{n-1}$?
$endgroup$
– user638500
Jan 26 at 17:14
$begingroup$
$T_0$ is some start value which we don't know a priori, since we only identified differences. $T_{n-1}$ disappears by telescoping: $T_n = T_n - T_{n-1} + T_{n-1} - T_{n-2} + ... + T_{2} - T_{1} + T_0$. You regroup that to $T_n = (T_n - T_{n-1}) + (T_{n-1} - T_{n-2}) + ... + (T_{2} - T_{1}) + T_0$ which gives you the fomula for $T_n$.
$endgroup$
– Andreas
Jan 27 at 12:29
add a comment |
$begingroup$
Can you please explain what $T_0$ is and what happened to $T_{n-1}$?
$endgroup$
– user638500
Jan 26 at 17:14
$begingroup$
$T_0$ is some start value which we don't know a priori, since we only identified differences. $T_{n-1}$ disappears by telescoping: $T_n = T_n - T_{n-1} + T_{n-1} - T_{n-2} + ... + T_{2} - T_{1} + T_0$. You regroup that to $T_n = (T_n - T_{n-1}) + (T_{n-1} - T_{n-2}) + ... + (T_{2} - T_{1}) + T_0$ which gives you the fomula for $T_n$.
$endgroup$
– Andreas
Jan 27 at 12:29
$begingroup$
Can you please explain what $T_0$ is and what happened to $T_{n-1}$?
$endgroup$
– user638500
Jan 26 at 17:14
$begingroup$
Can you please explain what $T_0$ is and what happened to $T_{n-1}$?
$endgroup$
– user638500
Jan 26 at 17:14
$begingroup$
$T_0$ is some start value which we don't know a priori, since we only identified differences. $T_{n-1}$ disappears by telescoping: $T_n = T_n - T_{n-1} + T_{n-1} - T_{n-2} + ... + T_{2} - T_{1} + T_0$. You regroup that to $T_n = (T_n - T_{n-1}) + (T_{n-1} - T_{n-2}) + ... + (T_{2} - T_{1}) + T_0$ which gives you the fomula for $T_n$.
$endgroup$
– Andreas
Jan 27 at 12:29
$begingroup$
$T_0$ is some start value which we don't know a priori, since we only identified differences. $T_{n-1}$ disappears by telescoping: $T_n = T_n - T_{n-1} + T_{n-1} - T_{n-2} + ... + T_{2} - T_{1} + T_0$. You regroup that to $T_n = (T_n - T_{n-1}) + (T_{n-1} - T_{n-2}) + ... + (T_{2} - T_{1}) + T_0$ which gives you the fomula for $T_n$.
$endgroup$
– Andreas
Jan 27 at 12:29
add a comment |
$begingroup$
Now if you define $R_n=T_n+n$ you can get the recurrence $$R_n-R_{n-1}=6cdot 2^{n-2}\R_1=T_1-1=8$$
and you get an arithmetico-geometric series for $R$
answered Jan 26 at 16:06
Ross MillikanRoss Millikan
300k24200374
$endgroup$
add a comment |
$begingroup$
Now if you define $R_n=T_n+n$ you can get the recurrence $$R_n-R_{n-1}=6cdot 2^{n-2}\R_1=T_1-1=8$$
and you get an arithmetico-geometric series for $R$
answered Jan 26 at 16:06
Ross MillikanRoss Millikan
300k24200374
$endgroup$
add a comment |
$begingroup$
Now if you define $R_n=T_n+n$ you can get the recurrence $$R_n-R_{n-1}=6cdot 2^{n-2}\R_1=T_1-1=8$$
and you get an arithmetico-geometric series for $R$
answered Jan 26 at 16:06
Ross MillikanRoss Millikan
300k24200374
$endgroup$
Now if you define $R_n=T_n+n$ you can get the recurrence $$R_n-R_{n-1}=6cdot 2^{n-2}\R_1=T_1-1=8$$
and you get an arithmetico-geometric series for $R$
answered Jan 26 at 16:06
Ross MillikanRoss Millikan
300k24200374
answered Jan 26 at 16:06
Ross MillikanRoss Millikan
300k24200374
answered Jan 26 at 16:06
Ross MillikanRoss Millikan
300k24200374
answered Jan 26 at 16:06
Ross MillikanRoss Millikan
300k24200374
300k24200374
add a comment |
add a comment |
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