Proving $binom{n+1+m}{n+1}=sum_{k=0}^m(k+1)binom{n+m-k}{n}$
$begingroup$
Use any method to prove that
$$binom{n+1+m}{n+1}=sum_{k=0}^m(k+1)binom{n+m-k}{n}$$
My Try:
Base case: Let $m=1$
LHS$$binom{n+1+m}{n+1}=binom{n+2}{n+1}=(n+2)$$
RHS$$sum_{k=0}^m(k+1)binom{n+m-k}{n}=binom{n+1-0}{n}+(1+1)binom{n+1-1}{n}$$
$$=frac{(n+1)!}{n!}+2$$
$$=(n+3)$$
If $m=2$
LHS$$binom{n+3}{n+1}=frac{n^2+5n+6}{2!}$$
RHS$$=binom{n+2}{n}+2binom{n+1}{n}+3binom{n}{n}$$
$$=frac{n^2+3n+2+4n+4+6}{2}=frac{n^2+7n+12}{2}$$
Clearly $LHSne RHS$
If LHS and RHS are not equal then how to prove this proof? Can anyone explain how to prove this.
combinatorics discrete-mathematics induction combinatorial-proofs
asked Dec 8 '18 at 16:37
user982787user982787
1117
$endgroup$
add a comment |
$begingroup$
Use any method to prove that
$$binom{n+1+m}{n+1}=sum_{k=0}^m(k+1)binom{n+m-k}{n}$$
My Try:
Base case: Let $m=1$
LHS$$binom{n+1+m}{n+1}=binom{n+2}{n+1}=(n+2)$$
RHS$$sum_{k=0}^m(k+1)binom{n+m-k}{n}=binom{n+1-0}{n}+(1+1)binom{n+1-1}{n}$$
$$=frac{(n+1)!}{n!}+2$$
$$=(n+3)$$
If $m=2$
LHS$$binom{n+3}{n+1}=frac{n^2+5n+6}{2!}$$
RHS$$=binom{n+2}{n}+2binom{n+1}{n}+3binom{n}{n}$$
$$=frac{n^2+3n+2+4n+4+6}{2}=frac{n^2+7n+12}{2}$$
Clearly $LHSne RHS$
If LHS and RHS are not equal then how to prove this proof? Can anyone explain how to prove this.
combinatorics discrete-mathematics induction combinatorial-proofs
asked Dec 8 '18 at 16:37
user982787user982787
1117
$endgroup$
4
$begingroup$
If something is not true, you cannot prove that it is true.
$endgroup$
– Batominovski
Dec 8 '18 at 16:46
$begingroup$
@Batominovski If it is wrong, then why am I asked to prove the equation.
$endgroup$
– user982787
Dec 8 '18 at 17:23
1
$begingroup$
Mistakes happen. I hope you understand that humans are not perfect, regardless of their intelligence and experiences.
$endgroup$
– Batominovski
Dec 8 '18 at 17:26
add a comment |
$begingroup$
Use any method to prove that
$$binom{n+1+m}{n+1}=sum_{k=0}^m(k+1)binom{n+m-k}{n}$$
My Try:
Base case: Let $m=1$
LHS$$binom{n+1+m}{n+1}=binom{n+2}{n+1}=(n+2)$$
RHS$$sum_{k=0}^m(k+1)binom{n+m-k}{n}=binom{n+1-0}{n}+(1+1)binom{n+1-1}{n}$$
$$=frac{(n+1)!}{n!}+2$$
$$=(n+3)$$
If $m=2$
LHS$$binom{n+3}{n+1}=frac{n^2+5n+6}{2!}$$
RHS$$=binom{n+2}{n}+2binom{n+1}{n}+3binom{n}{n}$$
$$=frac{n^2+3n+2+4n+4+6}{2}=frac{n^2+7n+12}{2}$$
Clearly $LHSne RHS$
If LHS and RHS are not equal then how to prove this proof? Can anyone explain how to prove this.
combinatorics discrete-mathematics induction combinatorial-proofs
asked Dec 8 '18 at 16:37
user982787user982787
1117
$endgroup$
Use any method to prove that
$$binom{n+1+m}{n+1}=sum_{k=0}^m(k+1)binom{n+m-k}{n}$$
My Try:
Base case: Let $m=1$
LHS$$binom{n+1+m}{n+1}=binom{n+2}{n+1}=(n+2)$$
RHS$$sum_{k=0}^m(k+1)binom{n+m-k}{n}=binom{n+1-0}{n}+(1+1)binom{n+1-1}{n}$$
$$=frac{(n+1)!}{n!}+2$$
$$=(n+3)$$
If $m=2$
LHS$$binom{n+3}{n+1}=frac{n^2+5n+6}{2!}$$
RHS$$=binom{n+2}{n}+2binom{n+1}{n}+3binom{n}{n}$$
$$=frac{n^2+3n+2+4n+4+6}{2}=frac{n^2+7n+12}{2}$$
Clearly $LHSne RHS$
If LHS and RHS are not equal then how to prove this proof? Can anyone explain how to prove this.
combinatorics discrete-mathematics induction combinatorial-proofs
combinatorics discrete-mathematics induction combinatorial-proofs
asked Dec 8 '18 at 16:37
user982787user982787
1117
asked Dec 8 '18 at 16:37
user982787user982787
1117
asked Dec 8 '18 at 16:37
user982787user982787
1117
asked Dec 8 '18 at 16:37
user982787user982787
1117
asked Dec 8 '18 at 16:37
user982787user982787
1117
1117
4
$begingroup$
If something is not true, you cannot prove that it is true.
$endgroup$
– Batominovski
Dec 8 '18 at 16:46
$begingroup$
@Batominovski If it is wrong, then why am I asked to prove the equation.
$endgroup$
– user982787
Dec 8 '18 at 17:23
1
$begingroup$
Mistakes happen. I hope you understand that humans are not perfect, regardless of their intelligence and experiences.
$endgroup$
– Batominovski
Dec 8 '18 at 17:26
add a comment |
4
$begingroup$
If something is not true, you cannot prove that it is true.
$endgroup$
– Batominovski
Dec 8 '18 at 16:46
$begingroup$
@Batominovski If it is wrong, then why am I asked to prove the equation.
$endgroup$
– user982787
Dec 8 '18 at 17:23
1
$begingroup$
Mistakes happen. I hope you understand that humans are not perfect, regardless of their intelligence and experiences.
$endgroup$
– Batominovski
Dec 8 '18 at 17:26
4
4
$begingroup$
If something is not true, you cannot prove that it is true.
$endgroup$
– Batominovski
Dec 8 '18 at 16:46
$begingroup$
If something is not true, you cannot prove that it is true.
$endgroup$
– Batominovski
Dec 8 '18 at 16:46
$begingroup$
@Batominovski If it is wrong, then why am I asked to prove the equation.
$endgroup$
– user982787
Dec 8 '18 at 17:23
$begingroup$
@Batominovski If it is wrong, then why am I asked to prove the equation.
$endgroup$
– user982787
Dec 8 '18 at 17:23
1
1
$begingroup$
Mistakes happen. I hope you understand that humans are not perfect, regardless of their intelligence and experiences.
$endgroup$
– Batominovski
Dec 8 '18 at 17:26
$begingroup$
Mistakes happen. I hope you understand that humans are not perfect, regardless of their intelligence and experiences.
$endgroup$
– Batominovski
Dec 8 '18 at 17:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{require{cancel}bcancel{cancel{n + 1 + m choose n + 1}} =
sum_{k = 0}^{m}pars{k + 1}{n + m - k choose n}: {LARGE ?}}$.
The right answer is
$bbx{ds{n + m + 2 choose n + 2}}$
$$
bbx{mbox{Note that} {n + m - k choose n} = 0
mbox{when} k > m}
$$
begin{align}
&bbox[10px,#ffd]{sum_{k = 0}^{m}pars{k + 1}
{n + m - k choose n}} =
sum_{k = 0}^{infty}pars{k + 1}{n + m - k choose m - k}
\[5mm] = &
sum_{k = 0}^{infty}pars{k + 1}
bracks{{-n - 1 choose m - k}pars{-1}^{m - k}}
\[5mm] = &
pars{-1}^{m}sum_{k = 0}^{infty}pars{k + 1}pars{-1}^{k}
bracks{z^{m - k}}pars{1 + z}^{-n - 1}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1}
sum_{k = 0}^{infty}pars{k + 1}pars{-z}^{k}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1},
pars{-,partiald{}{z}sum_{k = 0}^{infty}pars{-z}^{k + 1}}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1},
partiald{}{z}pars{z over 1 + z} =
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 3}
\[5mm] = &
pars{-1}^{m}{-n - 3 choose m} =
pars{-1}^{m}bracks{{n + 3 + m - 1choose m}pars{-1}^{m}}
\[5mm] = &
bbx{n + m + 2 choose n + 2}
end{align}
answered Dec 8 '18 at 20:30
Felix MarinFelix Marin
67.9k7107142
$endgroup$
add a comment |
$begingroup$
The RHS can be written as $$sum_{i+j=n+m+1}binom{i}1binom{j}n$$where $binom{r}{s}:=0$ if $snotin{0,dots,r}$.
This equals: $$binom{n+2+m}{n+2}$$
See here for a proof of that. So RHS does not equal LHS.
answered Dec 8 '18 at 16:53
drhabdrhab
101k544130
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{require{cancel}bcancel{cancel{n + 1 + m choose n + 1}} =
sum_{k = 0}^{m}pars{k + 1}{n + m - k choose n}: {LARGE ?}}$.
The right answer is
$bbx{ds{n + m + 2 choose n + 2}}$
$$
bbx{mbox{Note that} {n + m - k choose n} = 0
mbox{when} k > m}
$$
begin{align}
&bbox[10px,#ffd]{sum_{k = 0}^{m}pars{k + 1}
{n + m - k choose n}} =
sum_{k = 0}^{infty}pars{k + 1}{n + m - k choose m - k}
\[5mm] = &
sum_{k = 0}^{infty}pars{k + 1}
bracks{{-n - 1 choose m - k}pars{-1}^{m - k}}
\[5mm] = &
pars{-1}^{m}sum_{k = 0}^{infty}pars{k + 1}pars{-1}^{k}
bracks{z^{m - k}}pars{1 + z}^{-n - 1}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1}
sum_{k = 0}^{infty}pars{k + 1}pars{-z}^{k}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1},
pars{-,partiald{}{z}sum_{k = 0}^{infty}pars{-z}^{k + 1}}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1},
partiald{}{z}pars{z over 1 + z} =
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 3}
\[5mm] = &
pars{-1}^{m}{-n - 3 choose m} =
pars{-1}^{m}bracks{{n + 3 + m - 1choose m}pars{-1}^{m}}
\[5mm] = &
bbx{n + m + 2 choose n + 2}
end{align}
answered Dec 8 '18 at 20:30
Felix MarinFelix Marin
67.9k7107142
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{require{cancel}bcancel{cancel{n + 1 + m choose n + 1}} =
sum_{k = 0}^{m}pars{k + 1}{n + m - k choose n}: {LARGE ?}}$.
The right answer is
$bbx{ds{n + m + 2 choose n + 2}}$
$$
bbx{mbox{Note that} {n + m - k choose n} = 0
mbox{when} k > m}
$$
begin{align}
&bbox[10px,#ffd]{sum_{k = 0}^{m}pars{k + 1}
{n + m - k choose n}} =
sum_{k = 0}^{infty}pars{k + 1}{n + m - k choose m - k}
\[5mm] = &
sum_{k = 0}^{infty}pars{k + 1}
bracks{{-n - 1 choose m - k}pars{-1}^{m - k}}
\[5mm] = &
pars{-1}^{m}sum_{k = 0}^{infty}pars{k + 1}pars{-1}^{k}
bracks{z^{m - k}}pars{1 + z}^{-n - 1}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1}
sum_{k = 0}^{infty}pars{k + 1}pars{-z}^{k}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1},
pars{-,partiald{}{z}sum_{k = 0}^{infty}pars{-z}^{k + 1}}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1},
partiald{}{z}pars{z over 1 + z} =
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 3}
\[5mm] = &
pars{-1}^{m}{-n - 3 choose m} =
pars{-1}^{m}bracks{{n + 3 + m - 1choose m}pars{-1}^{m}}
\[5mm] = &
bbx{n + m + 2 choose n + 2}
end{align}
answered Dec 8 '18 at 20:30
Felix MarinFelix Marin
67.9k7107142
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
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newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{require{cancel}bcancel{cancel{n + 1 + m choose n + 1}} =
sum_{k = 0}^{m}pars{k + 1}{n + m - k choose n}: {LARGE ?}}$.
The right answer is
$bbx{ds{n + m + 2 choose n + 2}}$
$$
bbx{mbox{Note that} {n + m - k choose n} = 0
mbox{when} k > m}
$$
begin{align}
&bbox[10px,#ffd]{sum_{k = 0}^{m}pars{k + 1}
{n + m - k choose n}} =
sum_{k = 0}^{infty}pars{k + 1}{n + m - k choose m - k}
\[5mm] = &
sum_{k = 0}^{infty}pars{k + 1}
bracks{{-n - 1 choose m - k}pars{-1}^{m - k}}
\[5mm] = &
pars{-1}^{m}sum_{k = 0}^{infty}pars{k + 1}pars{-1}^{k}
bracks{z^{m - k}}pars{1 + z}^{-n - 1}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1}
sum_{k = 0}^{infty}pars{k + 1}pars{-z}^{k}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1},
pars{-,partiald{}{z}sum_{k = 0}^{infty}pars{-z}^{k + 1}}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1},
partiald{}{z}pars{z over 1 + z} =
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 3}
\[5mm] = &
pars{-1}^{m}{-n - 3 choose m} =
pars{-1}^{m}bracks{{n + 3 + m - 1choose m}pars{-1}^{m}}
\[5mm] = &
bbx{n + m + 2 choose n + 2}
end{align}
answered Dec 8 '18 at 20:30
Felix MarinFelix Marin
67.9k7107142
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{require{cancel}bcancel{cancel{n + 1 + m choose n + 1}} =
sum_{k = 0}^{m}pars{k + 1}{n + m - k choose n}: {LARGE ?}}$.
The right answer is
$bbx{ds{n + m + 2 choose n + 2}}$
$$
bbx{mbox{Note that} {n + m - k choose n} = 0
mbox{when} k > m}
$$
begin{align}
&bbox[10px,#ffd]{sum_{k = 0}^{m}pars{k + 1}
{n + m - k choose n}} =
sum_{k = 0}^{infty}pars{k + 1}{n + m - k choose m - k}
\[5mm] = &
sum_{k = 0}^{infty}pars{k + 1}
bracks{{-n - 1 choose m - k}pars{-1}^{m - k}}
\[5mm] = &
pars{-1}^{m}sum_{k = 0}^{infty}pars{k + 1}pars{-1}^{k}
bracks{z^{m - k}}pars{1 + z}^{-n - 1}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1}
sum_{k = 0}^{infty}pars{k + 1}pars{-z}^{k}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1},
pars{-,partiald{}{z}sum_{k = 0}^{infty}pars{-z}^{k + 1}}
\[5mm] = &
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 1},
partiald{}{z}pars{z over 1 + z} =
pars{-1}^{m}bracks{z^{m}}pars{1 + z}^{-n - 3}
\[5mm] = &
pars{-1}^{m}{-n - 3 choose m} =
pars{-1}^{m}bracks{{n + 3 + m - 1choose m}pars{-1}^{m}}
\[5mm] = &
bbx{n + m + 2 choose n + 2}
end{align}
answered Dec 8 '18 at 20:30
Felix MarinFelix Marin
67.9k7107142
answered Dec 8 '18 at 20:30
Felix MarinFelix Marin
67.9k7107142
answered Dec 8 '18 at 20:30
Felix MarinFelix Marin
67.9k7107142
answered Dec 8 '18 at 20:30
Felix MarinFelix Marin
67.9k7107142
67.9k7107142
add a comment |
add a comment |
$begingroup$
The RHS can be written as $$sum_{i+j=n+m+1}binom{i}1binom{j}n$$where $binom{r}{s}:=0$ if $snotin{0,dots,r}$.
This equals: $$binom{n+2+m}{n+2}$$
See here for a proof of that. So RHS does not equal LHS.
answered Dec 8 '18 at 16:53
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
The RHS can be written as $$sum_{i+j=n+m+1}binom{i}1binom{j}n$$where $binom{r}{s}:=0$ if $snotin{0,dots,r}$.
This equals: $$binom{n+2+m}{n+2}$$
See here for a proof of that. So RHS does not equal LHS.
answered Dec 8 '18 at 16:53
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
The RHS can be written as $$sum_{i+j=n+m+1}binom{i}1binom{j}n$$where $binom{r}{s}:=0$ if $snotin{0,dots,r}$.
This equals: $$binom{n+2+m}{n+2}$$
See here for a proof of that. So RHS does not equal LHS.
answered Dec 8 '18 at 16:53
drhabdrhab
101k544130
$endgroup$
The RHS can be written as $$sum_{i+j=n+m+1}binom{i}1binom{j}n$$where $binom{r}{s}:=0$ if $snotin{0,dots,r}$.
This equals: $$binom{n+2+m}{n+2}$$
See here for a proof of that. So RHS does not equal LHS.
answered Dec 8 '18 at 16:53
drhabdrhab
101k544130
edited Dec 8 '18 at 17:01
answered Dec 8 '18 at 16:53
drhabdrhab
101k544130
answered Dec 8 '18 at 16:53
drhabdrhab
101k544130
answered Dec 8 '18 at 16:53
drhabdrhab
101k544130
101k544130
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4
$begingroup$
If something is not true, you cannot prove that it is true.
$endgroup$
– Batominovski
Dec 8 '18 at 16:46
$begingroup$
@Batominovski If it is wrong, then why am I asked to prove the equation.
$endgroup$
– user982787
Dec 8 '18 at 17:23
1
$begingroup$
Mistakes happen. I hope you understand that humans are not perfect, regardless of their intelligence and experiences.
$endgroup$
– Batominovski
Dec 8 '18 at 17:26