What will be the sum of the numbers in the $100^{text{th}}$ step?
So this question states that a triangle is found where in between every $2$ numbers, their sum will be found in the next step, an example can be found below for the first $3$ steps. The question asks what will be the sum of the numbers in the $100^{text{th}}$ step? I managed to find a pattern that states $S_n=S_{n-1}+2 times 3^{n-1}$ if $S_k$ is the sum of step $k$. However, is this pattern correct and if it is then why? Also, can this triangle relate to Pascal's triangle? Thank you anyways.
combinatorics
asked Dec 2 '18 at 14:31
user587054
45711
|
show 2 more comments
So this question states that a triangle is found where in between every $2$ numbers, their sum will be found in the next step, an example can be found below for the first $3$ steps. The question asks what will be the sum of the numbers in the $100^{text{th}}$ step? I managed to find a pattern that states $S_n=S_{n-1}+2 times 3^{n-1}$ if $S_k$ is the sum of step $k$. However, is this pattern correct and if it is then why? Also, can this triangle relate to Pascal's triangle? Thank you anyways.
combinatorics
asked Dec 2 '18 at 14:31
user587054
45711
2
But the third row adds to $28$, your formula gives $26$. Can you write one more row?
– farruhota
Dec 2 '18 at 14:42
@Krishna I noticed that this triangle is similar to Pascal's and that is how I got the $2^n$ in my equation. However, the second part of my equation would be the harder one to prove as it does not have a huge relation with Pascal's triangle.
– user587054
Dec 2 '18 at 14:42
@farruhota Thank you for that notice I realized I had a mistake and meant to say $S_n=S_{n-1}+2 times 3^{n-1}$ This will be fixed.
– user587054
Dec 2 '18 at 14:47
I do not understand how the third row is made from the second, let alone how subsequent rows are made.
– Servaes
Dec 2 '18 at 14:47
@Servaes The same numbers from the second row have been written down again but this time with an extra number between each $2$ numbers being their sum. So the $4$ came from being the sum of $1+3$ and the from the $5$ sum of $3+2$
– user587054
Dec 2 '18 at 14:52
|
show 2 more comments
So this question states that a triangle is found where in between every $2$ numbers, their sum will be found in the next step, an example can be found below for the first $3$ steps. The question asks what will be the sum of the numbers in the $100^{text{th}}$ step? I managed to find a pattern that states $S_n=S_{n-1}+2 times 3^{n-1}$ if $S_k$ is the sum of step $k$. However, is this pattern correct and if it is then why? Also, can this triangle relate to Pascal's triangle? Thank you anyways.
combinatorics
asked Dec 2 '18 at 14:31
user587054
45711
So this question states that a triangle is found where in between every $2$ numbers, their sum will be found in the next step, an example can be found below for the first $3$ steps. The question asks what will be the sum of the numbers in the $100^{text{th}}$ step? I managed to find a pattern that states $S_n=S_{n-1}+2 times 3^{n-1}$ if $S_k$ is the sum of step $k$. However, is this pattern correct and if it is then why? Also, can this triangle relate to Pascal's triangle? Thank you anyways.
combinatorics
combinatorics
asked Dec 2 '18 at 14:31
user587054
45711
asked Dec 2 '18 at 14:31
user587054
45711
edited Dec 2 '18 at 14:48
asked Dec 2 '18 at 14:31
user587054
45711
asked Dec 2 '18 at 14:31
user587054
45711
asked Dec 2 '18 at 14:31
user587054
45711
45711
2
But the third row adds to $28$, your formula gives $26$. Can you write one more row?
– farruhota
Dec 2 '18 at 14:42
@Krishna I noticed that this triangle is similar to Pascal's and that is how I got the $2^n$ in my equation. However, the second part of my equation would be the harder one to prove as it does not have a huge relation with Pascal's triangle.
– user587054
Dec 2 '18 at 14:42
@farruhota Thank you for that notice I realized I had a mistake and meant to say $S_n=S_{n-1}+2 times 3^{n-1}$ This will be fixed.
– user587054
Dec 2 '18 at 14:47
I do not understand how the third row is made from the second, let alone how subsequent rows are made.
– Servaes
Dec 2 '18 at 14:47
@Servaes The same numbers from the second row have been written down again but this time with an extra number between each $2$ numbers being their sum. So the $4$ came from being the sum of $1+3$ and the from the $5$ sum of $3+2$
– user587054
Dec 2 '18 at 14:52
|
show 2 more comments
2
But the third row adds to $28$, your formula gives $26$. Can you write one more row?
– farruhota
Dec 2 '18 at 14:42
@Krishna I noticed that this triangle is similar to Pascal's and that is how I got the $2^n$ in my equation. However, the second part of my equation would be the harder one to prove as it does not have a huge relation with Pascal's triangle.
– user587054
Dec 2 '18 at 14:42
@farruhota Thank you for that notice I realized I had a mistake and meant to say $S_n=S_{n-1}+2 times 3^{n-1}$ This will be fixed.
– user587054
Dec 2 '18 at 14:47
I do not understand how the third row is made from the second, let alone how subsequent rows are made.
– Servaes
Dec 2 '18 at 14:47
@Servaes The same numbers from the second row have been written down again but this time with an extra number between each $2$ numbers being their sum. So the $4$ came from being the sum of $1+3$ and the from the $5$ sum of $3+2$
– user587054
Dec 2 '18 at 14:52
2
2
But the third row adds to $28$, your formula gives $26$. Can you write one more row?
– farruhota
Dec 2 '18 at 14:42
But the third row adds to $28$, your formula gives $26$. Can you write one more row?
– farruhota
Dec 2 '18 at 14:42
@Krishna I noticed that this triangle is similar to Pascal's and that is how I got the $2^n$ in my equation. However, the second part of my equation would be the harder one to prove as it does not have a huge relation with Pascal's triangle.
– user587054
Dec 2 '18 at 14:42
@Krishna I noticed that this triangle is similar to Pascal's and that is how I got the $2^n$ in my equation. However, the second part of my equation would be the harder one to prove as it does not have a huge relation with Pascal's triangle.
– user587054
Dec 2 '18 at 14:42
@farruhota Thank you for that notice I realized I had a mistake and meant to say $S_n=S_{n-1}+2 times 3^{n-1}$ This will be fixed.
– user587054
Dec 2 '18 at 14:47
@farruhota Thank you for that notice I realized I had a mistake and meant to say $S_n=S_{n-1}+2 times 3^{n-1}$ This will be fixed.
– user587054
Dec 2 '18 at 14:47
I do not understand how the third row is made from the second, let alone how subsequent rows are made.
– Servaes
Dec 2 '18 at 14:47
I do not understand how the third row is made from the second, let alone how subsequent rows are made.
– Servaes
Dec 2 '18 at 14:47
@Servaes The same numbers from the second row have been written down again but this time with an extra number between each $2$ numbers being their sum. So the $4$ came from being the sum of $1+3$ and the from the $5$ sum of $3+2$
– user587054
Dec 2 '18 at 14:52
@Servaes The same numbers from the second row have been written down again but this time with an extra number between each $2$ numbers being their sum. So the $4$ came from being the sum of $1+3$ and the from the $5$ sum of $3+2$
– user587054
Dec 2 '18 at 14:52
|
show 2 more comments
2 Answers
2
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Every number in the $n+1$-th row either comes directly from the $n$-th row, or is the sum of two neighbouring numbers in the $n$-th row. Every number in the $n$-th row has two neighbours (except the first and last) and so it is in two sums. So in summing the $n+1$-th row, we in fact sum every number from the $n$-th row three times (except the first and last). This yields the recursive formula
$$S_{n+1}=3S_n-2.$$
answered Dec 2 '18 at 14:53
Servaes
22.4k33793
add a comment |
Your recurrence equation is also correct. Note that:
$$S_n=S_{n-1}+2cdot 3^{n-1}, S_1=4 Rightarrow frac{S_n}{3^n}=frac13cdot frac{S_{n-1}}{3^{n-1}}+frac23 Rightarrow \
a_n=frac13a_{n-1}+frac23, a_1=frac43 Rightarrow a_n=left(frac13right)^n+1=frac{S_n}{3^n} Rightarrow \
S_n=3^n+1.$$
answered Dec 2 '18 at 15:51
farruhota
19.4k2736
add a comment |
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2 Answers
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2 Answers
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Every number in the $n+1$-th row either comes directly from the $n$-th row, or is the sum of two neighbouring numbers in the $n$-th row. Every number in the $n$-th row has two neighbours (except the first and last) and so it is in two sums. So in summing the $n+1$-th row, we in fact sum every number from the $n$-th row three times (except the first and last). This yields the recursive formula
$$S_{n+1}=3S_n-2.$$
answered Dec 2 '18 at 14:53
Servaes
22.4k33793
add a comment |
Every number in the $n+1$-th row either comes directly from the $n$-th row, or is the sum of two neighbouring numbers in the $n$-th row. Every number in the $n$-th row has two neighbours (except the first and last) and so it is in two sums. So in summing the $n+1$-th row, we in fact sum every number from the $n$-th row three times (except the first and last). This yields the recursive formula
$$S_{n+1}=3S_n-2.$$
answered Dec 2 '18 at 14:53
Servaes
22.4k33793
add a comment |
Every number in the $n+1$-th row either comes directly from the $n$-th row, or is the sum of two neighbouring numbers in the $n$-th row. Every number in the $n$-th row has two neighbours (except the first and last) and so it is in two sums. So in summing the $n+1$-th row, we in fact sum every number from the $n$-th row three times (except the first and last). This yields the recursive formula
$$S_{n+1}=3S_n-2.$$
answered Dec 2 '18 at 14:53
Servaes
22.4k33793
Every number in the $n+1$-th row either comes directly from the $n$-th row, or is the sum of two neighbouring numbers in the $n$-th row. Every number in the $n$-th row has two neighbours (except the first and last) and so it is in two sums. So in summing the $n+1$-th row, we in fact sum every number from the $n$-th row three times (except the first and last). This yields the recursive formula
$$S_{n+1}=3S_n-2.$$
answered Dec 2 '18 at 14:53
Servaes
22.4k33793
answered Dec 2 '18 at 14:53
Servaes
22.4k33793
answered Dec 2 '18 at 14:53
Servaes
22.4k33793
answered Dec 2 '18 at 14:53
Servaes
22.4k33793
22.4k33793
add a comment |
add a comment |
Your recurrence equation is also correct. Note that:
$$S_n=S_{n-1}+2cdot 3^{n-1}, S_1=4 Rightarrow frac{S_n}{3^n}=frac13cdot frac{S_{n-1}}{3^{n-1}}+frac23 Rightarrow \
a_n=frac13a_{n-1}+frac23, a_1=frac43 Rightarrow a_n=left(frac13right)^n+1=frac{S_n}{3^n} Rightarrow \
S_n=3^n+1.$$
answered Dec 2 '18 at 15:51
farruhota
19.4k2736
add a comment |
Your recurrence equation is also correct. Note that:
$$S_n=S_{n-1}+2cdot 3^{n-1}, S_1=4 Rightarrow frac{S_n}{3^n}=frac13cdot frac{S_{n-1}}{3^{n-1}}+frac23 Rightarrow \
a_n=frac13a_{n-1}+frac23, a_1=frac43 Rightarrow a_n=left(frac13right)^n+1=frac{S_n}{3^n} Rightarrow \
S_n=3^n+1.$$
answered Dec 2 '18 at 15:51
farruhota
19.4k2736
add a comment |
Your recurrence equation is also correct. Note that:
$$S_n=S_{n-1}+2cdot 3^{n-1}, S_1=4 Rightarrow frac{S_n}{3^n}=frac13cdot frac{S_{n-1}}{3^{n-1}}+frac23 Rightarrow \
a_n=frac13a_{n-1}+frac23, a_1=frac43 Rightarrow a_n=left(frac13right)^n+1=frac{S_n}{3^n} Rightarrow \
S_n=3^n+1.$$
answered Dec 2 '18 at 15:51
farruhota
19.4k2736
Your recurrence equation is also correct. Note that:
$$S_n=S_{n-1}+2cdot 3^{n-1}, S_1=4 Rightarrow frac{S_n}{3^n}=frac13cdot frac{S_{n-1}}{3^{n-1}}+frac23 Rightarrow \
a_n=frac13a_{n-1}+frac23, a_1=frac43 Rightarrow a_n=left(frac13right)^n+1=frac{S_n}{3^n} Rightarrow \
S_n=3^n+1.$$
answered Dec 2 '18 at 15:51
farruhota
19.4k2736
answered Dec 2 '18 at 15:51
farruhota
19.4k2736
answered Dec 2 '18 at 15:51
farruhota
19.4k2736
answered Dec 2 '18 at 15:51
farruhota
19.4k2736
19.4k2736
add a comment |
add a comment |
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2
But the third row adds to $28$, your formula gives $26$. Can you write one more row?
– farruhota
Dec 2 '18 at 14:42
@Krishna I noticed that this triangle is similar to Pascal's and that is how I got the $2^n$ in my equation. However, the second part of my equation would be the harder one to prove as it does not have a huge relation with Pascal's triangle.
– user587054
Dec 2 '18 at 14:42
@farruhota Thank you for that notice I realized I had a mistake and meant to say $S_n=S_{n-1}+2 times 3^{n-1}$ This will be fixed.
– user587054
Dec 2 '18 at 14:47
I do not understand how the third row is made from the second, let alone how subsequent rows are made.
– Servaes
Dec 2 '18 at 14:47
@Servaes The same numbers from the second row have been written down again but this time with an extra number between each $2$ numbers being their sum. So the $4$ came from being the sum of $1+3$ and the from the $5$ sum of $3+2$
– user587054
Dec 2 '18 at 14:52