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Simple question about a formula for sums

inequality of sumsFormula of Squaring Sums / IntegralsSolve easy sums with Binomial CoefficientWhy is this nested sum formula trueUsing a visual “proof” to show that $sum_{n=1}^{infty} left(frac 34 right)^n =1$Does an induction argument work when $n rightarrow infty$?Difficult rearrangement of summation terms to find analytical expressions for cumulant functionsquestion about sequences and seriesCalculate powers of sumsSimple question about Riemann zeta function

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1

$begingroup$

Why is
$$
left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n}a_ia_j
$$
I guess it is fairly clear when you do it for $n=2$ for example, but I can't really proof it for all $n$ with induction.

calculus analysis summation

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edited 8 hours ago

saulspatz

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asked 8 hours ago

KingDingelingKingDingeling

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$endgroup$

add a comment | 

1

$begingroup$

Why is
$$
left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n}a_ia_j
$$
I guess it is fairly clear when you do it for $n=2$ for example, but I can't really proof it for all $n$ with induction.

calculus analysis summation

share|cite|improve this question

edited 8 hours ago

saulspatz

22.2k44 gold badges1616 silver badges3838 bronze badges

asked 8 hours ago

KingDingelingKingDingeling

4501111 bronze badges

$endgroup$

add a comment | 

$begingroup$

Why is
$$
left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n}a_ia_j
$$
I guess it is fairly clear when you do it for $n=2$ for example, but I can't really proof it for all $n$ with induction.

calculus analysis summation

share|cite|improve this question

edited 8 hours ago

saulspatz

22.2k44 gold badges1616 silver badges3838 bronze badges

asked 8 hours ago

KingDingelingKingDingeling

4501111 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

saulspatz

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asked 8 hours ago

KingDingelingKingDingeling

4501111 bronze badges

share|cite|improve this question

edited 8 hours ago

saulspatz

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asked 8 hours ago

KingDingelingKingDingeling

4501111 bronze badges

edited 8 hours ago

saulspatz

22.2k44 gold badges1616 silver badges3838 bronze badges

edited 8 hours ago

saulspatz

22.2k44 gold badges1616 silver badges3838 bronze badges

asked 8 hours ago

KingDingelingKingDingeling

4501111 bronze badges

asked 8 hours ago

KingDingelingKingDingeling

4501111 bronze badges

2 Answers
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2

$begingroup$

NB: I give a proof by induction in the following. It may be a good exercise in order to practice induction, but the best way to understand the formula really is Saulspatz' point of view.

We may prove this identity by induction on $n$. We will make use of the base case $n=2$, which you may check by yourself.

The heredity is given by the following. Let $n>2$ be given and assume that the property holds for $n-1$. Let $a_1,ldots, a_n$ be given real numbers, complex numbers, or elements of any commutative ring. We define $A_1 = a_1+ldots +a_{n-1}$ and $A_2 = a_n$. Using the base case $n=2$, we may write

$$left(sum_{i=1}^{n}a_iright)^2=(A_1+A_2)^2=A_1^2 + A_1A_2+A_2A_1+A_2^2$$

Now, the induction hypothesis gives $A_1^2=sum_{i=1}^{n-1}sum_{j=1}^{n-1}a_ia_j$, and $A_2$ is no other than $a_n^2$.

As for $A_2A_1$, it is $sum_{j=1}^{n-1}a_na_j$. We may see it as the term "$i=n$" in order to regroup it inside the sum defining $A_1^2$. At this point, we have

$$left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n-1}a_ia_j+A_1A_2+a_n^2$$

As you may know, because we are considering finite sums, we may invert the order of summations. Thus, this is also

$$left(sum_{i=1}^{n}a_iright)^2= sum_{j=1}^{n-1}sum_{i=1}^{n}a_ia_j+A_1A_2+a_n^2$$

But $A_1A_2$ is $sum_{i=1}^{n-1}a_ia_n=left(sum_{i=1}^{n}a_ia_nright)-a_n^2$. Again, we identify the term "$j=n$" so that we may eventually regroup all the terms

$$left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n}a_ia_j$$

which is our desired identity.

share|cite|improve this answer

answered 8 hours ago

SuzetSuzet

2,83266 silver badges2727 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

On the left-hand side we have $$(a_1+a_2+cdots+a_n)(a_1+a_2+cdots+a_n)$$ How do we multiply this out? Pick a term in the first factor and a term in the second factor and multiply them. Add up the products for all possible choices of two terms. That's what the right-hand side says.

share|cite|improve this answer

answered 8 hours ago

saulspatzsaulspatz

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add a comment | 

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2

$begingroup$

NB: I give a proof by induction in the following. It may be a good exercise in order to practice induction, but the best way to understand the formula really is Saulspatz' point of view.

We may prove this identity by induction on $n$. We will make use of the base case $n=2$, which you may check by yourself.

The heredity is given by the following. Let $n>2$ be given and assume that the property holds for $n-1$. Let $a_1,ldots, a_n$ be given real numbers, complex numbers, or elements of any commutative ring. We define $A_1 = a_1+ldots +a_{n-1}$ and $A_2 = a_n$. Using the base case $n=2$, we may write

$$left(sum_{i=1}^{n}a_iright)^2=(A_1+A_2)^2=A_1^2 + A_1A_2+A_2A_1+A_2^2$$

Now, the induction hypothesis gives $A_1^2=sum_{i=1}^{n-1}sum_{j=1}^{n-1}a_ia_j$, and $A_2$ is no other than $a_n^2$.

As for $A_2A_1$, it is $sum_{j=1}^{n-1}a_na_j$. We may see it as the term "$i=n$" in order to regroup it inside the sum defining $A_1^2$. At this point, we have

$$left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n-1}a_ia_j+A_1A_2+a_n^2$$

As you may know, because we are considering finite sums, we may invert the order of summations. Thus, this is also

$$left(sum_{i=1}^{n}a_iright)^2= sum_{j=1}^{n-1}sum_{i=1}^{n}a_ia_j+A_1A_2+a_n^2$$

But $A_1A_2$ is $sum_{i=1}^{n-1}a_ia_n=left(sum_{i=1}^{n}a_ia_nright)-a_n^2$. Again, we identify the term "$j=n$" so that we may eventually regroup all the terms

$$left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n}a_ia_j$$

which is our desired identity.

share|cite|improve this answer

answered 8 hours ago

SuzetSuzet

2,83266 silver badges2727 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

NB: I give a proof by induction in the following. It may be a good exercise in order to practice induction, but the best way to understand the formula really is Saulspatz' point of view.

We may prove this identity by induction on $n$. We will make use of the base case $n=2$, which you may check by yourself.

The heredity is given by the following. Let $n>2$ be given and assume that the property holds for $n-1$. Let $a_1,ldots, a_n$ be given real numbers, complex numbers, or elements of any commutative ring. We define $A_1 = a_1+ldots +a_{n-1}$ and $A_2 = a_n$. Using the base case $n=2$, we may write

$$left(sum_{i=1}^{n}a_iright)^2=(A_1+A_2)^2=A_1^2 + A_1A_2+A_2A_1+A_2^2$$

Now, the induction hypothesis gives $A_1^2=sum_{i=1}^{n-1}sum_{j=1}^{n-1}a_ia_j$, and $A_2$ is no other than $a_n^2$.

As for $A_2A_1$, it is $sum_{j=1}^{n-1}a_na_j$. We may see it as the term "$i=n$" in order to regroup it inside the sum defining $A_1^2$. At this point, we have

$$left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n-1}a_ia_j+A_1A_2+a_n^2$$

As you may know, because we are considering finite sums, we may invert the order of summations. Thus, this is also

$$left(sum_{i=1}^{n}a_iright)^2= sum_{j=1}^{n-1}sum_{i=1}^{n}a_ia_j+A_1A_2+a_n^2$$

But $A_1A_2$ is $sum_{i=1}^{n-1}a_ia_n=left(sum_{i=1}^{n}a_ia_nright)-a_n^2$. Again, we identify the term "$j=n$" so that we may eventually regroup all the terms

$$left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n}a_ia_j$$

which is our desired identity.

share|cite|improve this answer

answered 8 hours ago

SuzetSuzet

2,83266 silver badges2727 bronze badges

$endgroup$

add a comment | 

$begingroup$

NB: I give a proof by induction in the following. It may be a good exercise in order to practice induction, but the best way to understand the formula really is Saulspatz' point of view.

We may prove this identity by induction on $n$. We will make use of the base case $n=2$, which you may check by yourself.

The heredity is given by the following. Let $n>2$ be given and assume that the property holds for $n-1$. Let $a_1,ldots, a_n$ be given real numbers, complex numbers, or elements of any commutative ring. We define $A_1 = a_1+ldots +a_{n-1}$ and $A_2 = a_n$. Using the base case $n=2$, we may write

$$left(sum_{i=1}^{n}a_iright)^2=(A_1+A_2)^2=A_1^2 + A_1A_2+A_2A_1+A_2^2$$

Now, the induction hypothesis gives $A_1^2=sum_{i=1}^{n-1}sum_{j=1}^{n-1}a_ia_j$, and $A_2$ is no other than $a_n^2$.

As for $A_2A_1$, it is $sum_{j=1}^{n-1}a_na_j$. We may see it as the term "$i=n$" in order to regroup it inside the sum defining $A_1^2$. At this point, we have

$$left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n-1}a_ia_j+A_1A_2+a_n^2$$

As you may know, because we are considering finite sums, we may invert the order of summations. Thus, this is also

$$left(sum_{i=1}^{n}a_iright)^2= sum_{j=1}^{n-1}sum_{i=1}^{n}a_ia_j+A_1A_2+a_n^2$$

But $A_1A_2$ is $sum_{i=1}^{n-1}a_ia_n=left(sum_{i=1}^{n}a_ia_nright)-a_n^2$. Again, we identify the term "$j=n$" so that we may eventually regroup all the terms

$$left(sum_{i=1}^{n}a_iright)^2= sum_{i=1}^{n}sum_{j=1}^{n}a_ia_j$$

which is our desired identity.

share|cite|improve this answer

answered 8 hours ago

SuzetSuzet

2,83266 silver badges2727 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

SuzetSuzet

2,83266 silver badges2727 bronze badges

answered 8 hours ago

SuzetSuzet

2,83266 silver badges2727 bronze badges

answered 8 hours ago

SuzetSuzet

2,83266 silver badges2727 bronze badges

4

$begingroup$

On the left-hand side we have $$(a_1+a_2+cdots+a_n)(a_1+a_2+cdots+a_n)$$ How do we multiply this out? Pick a term in the first factor and a term in the second factor and multiply them. Add up the products for all possible choices of two terms. That's what the right-hand side says.

share|cite|improve this answer

answered 8 hours ago

saulspatzsaulspatz

22.2k44 gold badges1616 silver badges3838 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

On the left-hand side we have $$(a_1+a_2+cdots+a_n)(a_1+a_2+cdots+a_n)$$ How do we multiply this out? Pick a term in the first factor and a term in the second factor and multiply them. Add up the products for all possible choices of two terms. That's what the right-hand side says.

share|cite|improve this answer

answered 8 hours ago

saulspatzsaulspatz

22.2k44 gold badges1616 silver badges3838 bronze badges

$endgroup$

add a comment | 

$begingroup$

On the left-hand side we have $$(a_1+a_2+cdots+a_n)(a_1+a_2+cdots+a_n)$$ How do we multiply this out? Pick a term in the first factor and a term in the second factor and multiply them. Add up the products for all possible choices of two terms. That's what the right-hand side says.

share|cite|improve this answer

answered 8 hours ago

saulspatzsaulspatz

22.2k44 gold badges1616 silver badges3838 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

saulspatzsaulspatz

22.2k44 gold badges1616 silver badges3838 bronze badges

answered 8 hours ago

saulspatzsaulspatz

22.2k44 gold badges1616 silver badges3838 bronze badges

answered 8 hours ago

saulspatzsaulspatz

22.2k44 gold badges1616 silver badges3838 bronze badges