Don't Understand Double Summation with Gauss's formula
So as we know, if we have a summation from $1$ to $n$, the simple formula is
$$sum_{i=1}^n i=frac{n(n+1)}2.$$
But if we have two summations, one from $i=1$ to $n$ and another one $j>i$ to $n$, the formula we get is
$$sum_{i=1}^nsum_{j>i}^nj=frac{n(n-1)}2.$$
I'm not understanding this, how does it go from $n+1$ to $n-1$, especially since it's a double summation?
summation
edited Nov 19 '18 at 3:10
Tianlalu
3,09621038
asked Nov 19 '18 at 3:05
reVolutionary
82
add a comment |
So as we know, if we have a summation from $1$ to $n$, the simple formula is
$$sum_{i=1}^n i=frac{n(n+1)}2.$$
But if we have two summations, one from $i=1$ to $n$ and another one $j>i$ to $n$, the formula we get is
$$sum_{i=1}^nsum_{j>i}^nj=frac{n(n-1)}2.$$
I'm not understanding this, how does it go from $n+1$ to $n-1$, especially since it's a double summation?
summation
edited Nov 19 '18 at 3:10
Tianlalu
3,09621038
asked Nov 19 '18 at 3:05
reVolutionary
82
I have edited your question. You can re-edit it if my edits are not correct.
– Tianlalu
Nov 19 '18 at 3:11
add a comment |
So as we know, if we have a summation from $1$ to $n$, the simple formula is
$$sum_{i=1}^n i=frac{n(n+1)}2.$$
But if we have two summations, one from $i=1$ to $n$ and another one $j>i$ to $n$, the formula we get is
$$sum_{i=1}^nsum_{j>i}^nj=frac{n(n-1)}2.$$
I'm not understanding this, how does it go from $n+1$ to $n-1$, especially since it's a double summation?
summation
edited Nov 19 '18 at 3:10
Tianlalu
3,09621038
asked Nov 19 '18 at 3:05
reVolutionary
82
So as we know, if we have a summation from $1$ to $n$, the simple formula is
$$sum_{i=1}^n i=frac{n(n+1)}2.$$
But if we have two summations, one from $i=1$ to $n$ and another one $j>i$ to $n$, the formula we get is
$$sum_{i=1}^nsum_{j>i}^nj=frac{n(n-1)}2.$$
I'm not understanding this, how does it go from $n+1$ to $n-1$, especially since it's a double summation?
summation
summation
edited Nov 19 '18 at 3:10
Tianlalu
3,09621038
asked Nov 19 '18 at 3:05
reVolutionary
82
edited Nov 19 '18 at 3:10
Tianlalu
3,09621038
asked Nov 19 '18 at 3:05
reVolutionary
82
edited Nov 19 '18 at 3:10
Tianlalu
3,09621038
edited Nov 19 '18 at 3:10
Tianlalu
3,09621038
edited Nov 19 '18 at 3:10
Tianlalu
3,09621038
3,09621038
asked Nov 19 '18 at 3:05
reVolutionary
82
asked Nov 19 '18 at 3:05
reVolutionary
82
asked Nov 19 '18 at 3:05
reVolutionary
82
82
I have edited your question. You can re-edit it if my edits are not correct.
– Tianlalu
Nov 19 '18 at 3:11
add a comment |
I have edited your question. You can re-edit it if my edits are not correct.
– Tianlalu
Nov 19 '18 at 3:11
I have edited your question. You can re-edit it if my edits are not correct.
– Tianlalu
Nov 19 '18 at 3:11
I have edited your question. You can re-edit it if my edits are not correct.
– Tianlalu
Nov 19 '18 at 3:11
add a comment |
1 Answer
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I think you might be confusing two different things.
Fact 1: Define $S_n := sum_{k=1}^n k$. Then $S_n = n(n+1)/2$ for every nonnegative integer $n$.
(You can see this by induction. Certainly the formula is correct for $n = 0$. With $S_{n-1} = (n-1)n/2$, we have $S_{n} = n + S_{n-1} = (n+1)n/2$, as required.)
Then there is the following fact
Fact 2: Define $T_n = sum_{1 leq i < j leq n} 1$. Then $T_n = n(n-1)/2$.
This is just the number of tuples $(k, l)$ with $1 leq k leq n$ and $1 leq l leq n$ there are with $k < l$. That's just $binom{n}{2}$, or if you like it's also $(n^2 - n)/2 = n(n-1)/2$.
answered Nov 19 '18 at 3:18
Drew Brady
649315
add a comment |
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1 Answer
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1 Answer
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oldest
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I think you might be confusing two different things.
Fact 1: Define $S_n := sum_{k=1}^n k$. Then $S_n = n(n+1)/2$ for every nonnegative integer $n$.
(You can see this by induction. Certainly the formula is correct for $n = 0$. With $S_{n-1} = (n-1)n/2$, we have $S_{n} = n + S_{n-1} = (n+1)n/2$, as required.)
Then there is the following fact
Fact 2: Define $T_n = sum_{1 leq i < j leq n} 1$. Then $T_n = n(n-1)/2$.
This is just the number of tuples $(k, l)$ with $1 leq k leq n$ and $1 leq l leq n$ there are with $k < l$. That's just $binom{n}{2}$, or if you like it's also $(n^2 - n)/2 = n(n-1)/2$.
answered Nov 19 '18 at 3:18
Drew Brady
649315
add a comment |
I think you might be confusing two different things.
Fact 1: Define $S_n := sum_{k=1}^n k$. Then $S_n = n(n+1)/2$ for every nonnegative integer $n$.
(You can see this by induction. Certainly the formula is correct for $n = 0$. With $S_{n-1} = (n-1)n/2$, we have $S_{n} = n + S_{n-1} = (n+1)n/2$, as required.)
Then there is the following fact
Fact 2: Define $T_n = sum_{1 leq i < j leq n} 1$. Then $T_n = n(n-1)/2$.
This is just the number of tuples $(k, l)$ with $1 leq k leq n$ and $1 leq l leq n$ there are with $k < l$. That's just $binom{n}{2}$, or if you like it's also $(n^2 - n)/2 = n(n-1)/2$.
answered Nov 19 '18 at 3:18
Drew Brady
649315
add a comment |
I think you might be confusing two different things.
Fact 1: Define $S_n := sum_{k=1}^n k$. Then $S_n = n(n+1)/2$ for every nonnegative integer $n$.
(You can see this by induction. Certainly the formula is correct for $n = 0$. With $S_{n-1} = (n-1)n/2$, we have $S_{n} = n + S_{n-1} = (n+1)n/2$, as required.)
Then there is the following fact
Fact 2: Define $T_n = sum_{1 leq i < j leq n} 1$. Then $T_n = n(n-1)/2$.
This is just the number of tuples $(k, l)$ with $1 leq k leq n$ and $1 leq l leq n$ there are with $k < l$. That's just $binom{n}{2}$, or if you like it's also $(n^2 - n)/2 = n(n-1)/2$.
answered Nov 19 '18 at 3:18
Drew Brady
649315
I think you might be confusing two different things.
Fact 1: Define $S_n := sum_{k=1}^n k$. Then $S_n = n(n+1)/2$ for every nonnegative integer $n$.
(You can see this by induction. Certainly the formula is correct for $n = 0$. With $S_{n-1} = (n-1)n/2$, we have $S_{n} = n + S_{n-1} = (n+1)n/2$, as required.)
Then there is the following fact
Fact 2: Define $T_n = sum_{1 leq i < j leq n} 1$. Then $T_n = n(n-1)/2$.
This is just the number of tuples $(k, l)$ with $1 leq k leq n$ and $1 leq l leq n$ there are with $k < l$. That's just $binom{n}{2}$, or if you like it's also $(n^2 - n)/2 = n(n-1)/2$.
answered Nov 19 '18 at 3:18
Drew Brady
649315
answered Nov 19 '18 at 3:18
Drew Brady
649315
answered Nov 19 '18 at 3:18
Drew Brady
649315
answered Nov 19 '18 at 3:18
Drew Brady
649315
649315
add a comment |
add a comment |
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I have edited your question. You can re-edit it if my edits are not correct.
– Tianlalu
Nov 19 '18 at 3:11