Property of summation
$begingroup$
Very short question. Could you please explain me why
$$sum_{i=0}^{n-1} a = na$$
with $a$ a constant?
I know that
$$sum_{i=1}^{n} a = na$$
but in my case the sum starts from zero and finishes for $(n-1)$.
Thanks.
summation
edited Mar 15 at 15:04
MarianD
1,4941616
asked Mar 15 at 14:26
KolmogorovwannabeKolmogorovwannabe
257
$endgroup$
add a comment |
$begingroup$
Very short question. Could you please explain me why
$$sum_{i=0}^{n-1} a = na$$
with $a$ a constant?
I know that
$$sum_{i=1}^{n} a = na$$
but in my case the sum starts from zero and finishes for $(n-1)$.
Thanks.
summation
edited Mar 15 at 15:04
MarianD
1,4941616
asked Mar 15 at 14:26
KolmogorovwannabeKolmogorovwannabe
257
$endgroup$
7
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
Mar 15 at 14:29
$begingroup$
Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
Mar 15 at 14:29
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
Mar 15 at 14:40
add a comment |
$begingroup$
Very short question. Could you please explain me why
$$sum_{i=0}^{n-1} a = na$$
with $a$ a constant?
I know that
$$sum_{i=1}^{n} a = na$$
but in my case the sum starts from zero and finishes for $(n-1)$.
Thanks.
summation
edited Mar 15 at 15:04
MarianD
1,4941616
asked Mar 15 at 14:26
KolmogorovwannabeKolmogorovwannabe
257
$endgroup$
Very short question. Could you please explain me why
$$sum_{i=0}^{n-1} a = na$$
with $a$ a constant?
I know that
$$sum_{i=1}^{n} a = na$$
but in my case the sum starts from zero and finishes for $(n-1)$.
Thanks.
summation
summation
edited Mar 15 at 15:04
MarianD
1,4941616
asked Mar 15 at 14:26
KolmogorovwannabeKolmogorovwannabe
257
edited Mar 15 at 15:04
MarianD
1,4941616
asked Mar 15 at 14:26
KolmogorovwannabeKolmogorovwannabe
257
edited Mar 15 at 15:04
MarianD
1,4941616
edited Mar 15 at 15:04
MarianD
1,4941616
edited Mar 15 at 15:04
MarianD
1,4941616
1,4941616
asked Mar 15 at 14:26
KolmogorovwannabeKolmogorovwannabe
257
asked Mar 15 at 14:26
KolmogorovwannabeKolmogorovwannabe
257
asked Mar 15 at 14:26
KolmogorovwannabeKolmogorovwannabe
257
257
7
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
Mar 15 at 14:29
$begingroup$
Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
Mar 15 at 14:29
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
Mar 15 at 14:40
add a comment |
7
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
Mar 15 at 14:29
$begingroup$
Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
Mar 15 at 14:29
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
Mar 15 at 14:40
7
7
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
Mar 15 at 14:29
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
Mar 15 at 14:29
$begingroup$
Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
Mar 15 at 14:29
$begingroup$
Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
Mar 15 at 14:29
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
Mar 15 at 14:40
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
Mar 15 at 14:40
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Since you are allready convinced that $sum_{i=1}^{n}a=na$ this might help:
$sum_{i=0}^{n-1}a=a+sum_{i=1}^{n-1}a=a+sum_{i=1}^{n}a-a=sum_{i=1}^{n}a$
answered Mar 15 at 14:30
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
In both cases - $$sum_{i=0}^{n-1}aquad text { and }quadsum_{i=1}^{n}a$$
- there are exactly $n$ summands.
answered Mar 15 at 15:07
MarianDMarianD
1,4941616
$endgroup$
add a comment |
$begingroup$
Since $a$ is not $i$-depending one can write: $$sum_{i=0}^{n-1}{a}=asum_{i=0}^{n-1}{1}$$
And $sum_{i=0}^{n-1}{1}=1+1+cdots+1$ $n$ times which obviously is $n$.
edited Mar 15 at 14:54
MarianD
1,4941616
answered Mar 15 at 14:33
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
$endgroup$
add a comment |
$begingroup$
What is the definition of $sum_{i=0}^{n-1} x_i$? It is exactly $x_0+x_1+...x_{n-1}$. If $x_0=x_1=...x_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
answered Mar 15 at 14:29
MarkMark
10.4k1622
$endgroup$
add a comment |
$begingroup$
You are summing $n$ terms, all equal to $a$. So
$$sum_{i=0}^{n-1} a = a+ a+dotsb + a = na.$$
answered Mar 15 at 14:29
HugoHugo
8206
$endgroup$
add a comment |
$begingroup$
Hint:
$a$ times the number of terms.
answered Mar 15 at 14:30
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since you are allready convinced that $sum_{i=1}^{n}a=na$ this might help:
$sum_{i=0}^{n-1}a=a+sum_{i=1}^{n-1}a=a+sum_{i=1}^{n}a-a=sum_{i=1}^{n}a$
answered Mar 15 at 14:30
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Since you are allready convinced that $sum_{i=1}^{n}a=na$ this might help:
$sum_{i=0}^{n-1}a=a+sum_{i=1}^{n-1}a=a+sum_{i=1}^{n}a-a=sum_{i=1}^{n}a$
answered Mar 15 at 14:30
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Since you are allready convinced that $sum_{i=1}^{n}a=na$ this might help:
$sum_{i=0}^{n-1}a=a+sum_{i=1}^{n-1}a=a+sum_{i=1}^{n}a-a=sum_{i=1}^{n}a$
answered Mar 15 at 14:30
drhabdrhab
103k545136
$endgroup$
Since you are allready convinced that $sum_{i=1}^{n}a=na$ this might help:
$sum_{i=0}^{n-1}a=a+sum_{i=1}^{n-1}a=a+sum_{i=1}^{n}a-a=sum_{i=1}^{n}a$
answered Mar 15 at 14:30
drhabdrhab
103k545136
answered Mar 15 at 14:30
drhabdrhab
103k545136
answered Mar 15 at 14:30
drhabdrhab
103k545136
answered Mar 15 at 14:30
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
$begingroup$
In both cases - $$sum_{i=0}^{n-1}aquad text { and }quadsum_{i=1}^{n}a$$
- there are exactly $n$ summands.
answered Mar 15 at 15:07
MarianDMarianD
1,4941616
$endgroup$
add a comment |
$begingroup$
In both cases - $$sum_{i=0}^{n-1}aquad text { and }quadsum_{i=1}^{n}a$$
- there are exactly $n$ summands.
answered Mar 15 at 15:07
MarianDMarianD
1,4941616
$endgroup$
add a comment |
$begingroup$
In both cases - $$sum_{i=0}^{n-1}aquad text { and }quadsum_{i=1}^{n}a$$
- there are exactly $n$ summands.
answered Mar 15 at 15:07
MarianDMarianD
1,4941616
$endgroup$
In both cases - $$sum_{i=0}^{n-1}aquad text { and }quadsum_{i=1}^{n}a$$
- there are exactly $n$ summands.
answered Mar 15 at 15:07
MarianDMarianD
1,4941616
answered Mar 15 at 15:07
MarianDMarianD
1,4941616
answered Mar 15 at 15:07
MarianDMarianD
1,4941616
answered Mar 15 at 15:07
MarianDMarianD
1,4941616
1,4941616
add a comment |
add a comment |
$begingroup$
Since $a$ is not $i$-depending one can write: $$sum_{i=0}^{n-1}{a}=asum_{i=0}^{n-1}{1}$$
And $sum_{i=0}^{n-1}{1}=1+1+cdots+1$ $n$ times which obviously is $n$.
edited Mar 15 at 14:54
MarianD
1,4941616
answered Mar 15 at 14:33
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
$endgroup$
add a comment |
$begingroup$
Since $a$ is not $i$-depending one can write: $$sum_{i=0}^{n-1}{a}=asum_{i=0}^{n-1}{1}$$
And $sum_{i=0}^{n-1}{1}=1+1+cdots+1$ $n$ times which obviously is $n$.
edited Mar 15 at 14:54
MarianD
1,4941616
answered Mar 15 at 14:33
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
$endgroup$
add a comment |
$begingroup$
Since $a$ is not $i$-depending one can write: $$sum_{i=0}^{n-1}{a}=asum_{i=0}^{n-1}{1}$$
And $sum_{i=0}^{n-1}{1}=1+1+cdots+1$ $n$ times which obviously is $n$.
edited Mar 15 at 14:54
MarianD
1,4941616
answered Mar 15 at 14:33
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
$endgroup$
Since $a$ is not $i$-depending one can write: $$sum_{i=0}^{n-1}{a}=asum_{i=0}^{n-1}{1}$$
And $sum_{i=0}^{n-1}{1}=1+1+cdots+1$ $n$ times which obviously is $n$.
edited Mar 15 at 14:54
MarianD
1,4941616
answered Mar 15 at 14:33
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
edited Mar 15 at 14:54
MarianD
1,4941616
edited Mar 15 at 14:54
MarianD
1,4941616
edited Mar 15 at 14:54
MarianD
1,4941616
1,4941616
answered Mar 15 at 14:33
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
answered Mar 15 at 14:33
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
answered Mar 15 at 14:33
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,20729
1,20729
add a comment |
add a comment |
$begingroup$
What is the definition of $sum_{i=0}^{n-1} x_i$? It is exactly $x_0+x_1+...x_{n-1}$. If $x_0=x_1=...x_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
answered Mar 15 at 14:29
MarkMark
10.4k1622
$endgroup$
add a comment |
$begingroup$
What is the definition of $sum_{i=0}^{n-1} x_i$? It is exactly $x_0+x_1+...x_{n-1}$. If $x_0=x_1=...x_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
answered Mar 15 at 14:29
MarkMark
10.4k1622
$endgroup$
add a comment |
$begingroup$
What is the definition of $sum_{i=0}^{n-1} x_i$? It is exactly $x_0+x_1+...x_{n-1}$. If $x_0=x_1=...x_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
answered Mar 15 at 14:29
MarkMark
10.4k1622
$endgroup$
What is the definition of $sum_{i=0}^{n-1} x_i$? It is exactly $x_0+x_1+...x_{n-1}$. If $x_0=x_1=...x_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
answered Mar 15 at 14:29
MarkMark
10.4k1622
edited Mar 15 at 15:33
answered Mar 15 at 14:29
MarkMark
10.4k1622
answered Mar 15 at 14:29
MarkMark
10.4k1622
answered Mar 15 at 14:29
MarkMark
10.4k1622
10.4k1622
add a comment |
add a comment |
$begingroup$
You are summing $n$ terms, all equal to $a$. So
$$sum_{i=0}^{n-1} a = a+ a+dotsb + a = na.$$
answered Mar 15 at 14:29
HugoHugo
8206
$endgroup$
add a comment |
$begingroup$
You are summing $n$ terms, all equal to $a$. So
$$sum_{i=0}^{n-1} a = a+ a+dotsb + a = na.$$
answered Mar 15 at 14:29
HugoHugo
8206
$endgroup$
add a comment |
$begingroup$
You are summing $n$ terms, all equal to $a$. So
$$sum_{i=0}^{n-1} a = a+ a+dotsb + a = na.$$
answered Mar 15 at 14:29
HugoHugo
8206
$endgroup$
You are summing $n$ terms, all equal to $a$. So
$$sum_{i=0}^{n-1} a = a+ a+dotsb + a = na.$$
answered Mar 15 at 14:29
HugoHugo
8206
answered Mar 15 at 14:29
HugoHugo
8206
answered Mar 15 at 14:29
HugoHugo
8206
answered Mar 15 at 14:29
HugoHugo
8206
8206
add a comment |
add a comment |
$begingroup$
Hint:
$a$ times the number of terms.
answered Mar 15 at 14:30
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
Hint:
$a$ times the number of terms.
answered Mar 15 at 14:30
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
Hint:
$a$ times the number of terms.
answered Mar 15 at 14:30
Yves DaoustYves Daoust
131k676229
$endgroup$
Hint:
$a$ times the number of terms.
answered Mar 15 at 14:30
Yves DaoustYves Daoust
131k676229
answered Mar 15 at 14:30
Yves DaoustYves Daoust
131k676229
answered Mar 15 at 14:30
Yves DaoustYves Daoust
131k676229
answered Mar 15 at 14:30
Yves DaoustYves Daoust
131k676229
131k676229
add a comment |
add a comment |
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7
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
Mar 15 at 14:29
$begingroup$
Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
Mar 15 at 14:29
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
Mar 15 at 14:40