Mixing
A double-sum identity
Clash Royale CLAN TAG#URR8PPP
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Let $a_k$ be a sequence. Now prove that
$$sum_j=0^n-1 frac12^j+1 sum_k=0^j(-1)^kbinomjka_k
=sum_k=0^n-1(-1)^ka_ksum_j=k^n-1 frac12^j+1 binomjk.$$
I have tried but I can't. I know the identity
$sum_k=0^n binomnk = 2^n$.
calculus summation-method
edited Jul 29 at 8:13
Robert Z
83.7k954122
asked Jul 29 at 7:42
Ole Petersen
1637
add a comment |Â
up vote
0
down vote
favorite
Let $a_k$ be a sequence. Now prove that
$$sum_j=0^n-1 frac12^j+1 sum_k=0^j(-1)^kbinomjka_k
=sum_k=0^n-1(-1)^ka_ksum_j=k^n-1 frac12^j+1 binomjk.$$
I have tried but I can't. I know the identity
$sum_k=0^n binomnk = 2^n$.
calculus summation-method
edited Jul 29 at 8:13
Robert Z
83.7k954122
asked Jul 29 at 7:42
Ole Petersen
1637
1
You are summing over the set of couples $(k,j): 0leq kleq jleq n-1$.
– Robert Z
Jul 29 at 7:48
How does this prove the above statement?
– Ole Petersen
Jul 29 at 7:52
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $a_k$ be a sequence. Now prove that
$$sum_j=0^n-1 frac12^j+1 sum_k=0^j(-1)^kbinomjka_k
=sum_k=0^n-1(-1)^ka_ksum_j=k^n-1 frac12^j+1 binomjk.$$
I have tried but I can't. I know the identity
$sum_k=0^n binomnk = 2^n$.
calculus summation-method
edited Jul 29 at 8:13
Robert Z
83.7k954122
asked Jul 29 at 7:42
Ole Petersen
1637
Let $a_k$ be a sequence. Now prove that
$$sum_j=0^n-1 frac12^j+1 sum_k=0^j(-1)^kbinomjka_k
=sum_k=0^n-1(-1)^ka_ksum_j=k^n-1 frac12^j+1 binomjk.$$
I have tried but I can't. I know the identity
$sum_k=0^n binomnk = 2^n$.
calculus summation-method
edited Jul 29 at 8:13
Robert Z
83.7k954122
asked Jul 29 at 7:42
Ole Petersen
1637
edited Jul 29 at 8:13
Robert Z
83.7k954122
edited Jul 29 at 8:13
Robert Z
83.7k954122
edited Jul 29 at 8:13
Robert Z
83.7k954122
83.7k954122
asked Jul 29 at 7:42
Ole Petersen
1637
asked Jul 29 at 7:42
Ole Petersen
1637
asked Jul 29 at 7:42
Ole Petersen
1637
1637
1
You are summing over the set of couples $(k,j): 0leq kleq jleq n-1$.
– Robert Z
Jul 29 at 7:48
How does this prove the above statement?
– Ole Petersen
Jul 29 at 7:52
add a comment |Â
1
You are summing over the set of couples $(k,j): 0leq kleq jleq n-1$.
– Robert Z
Jul 29 at 7:48
How does this prove the above statement?
– Ole Petersen
Jul 29 at 7:52
1
1
You are summing over the set of couples $(k,j): 0leq kleq jleq n-1$.
– Robert Z
Jul 29 at 7:48
You are summing over the set of couples $(k,j): 0leq kleq jleq n-1$.
– Robert Z
Jul 29 at 7:48
How does this prove the above statement?
– Ole Petersen
Jul 29 at 7:52
How does this prove the above statement?
– Ole Petersen
Jul 29 at 7:52
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
In the two given double sums, we are summing over the same set of couples $$S_n:=(j,k)inmathbbNtimes mathbbN: 0leq kleq jleq n-1.$$
Hence for ANY double sequence $(A(j,k))_j,k$,
$$sum_j=0^n-1sum_k=0^j A(j,k)=sum_(j,k)in S_nA(j,k)=
sum_k=0^n-1sum_j=k^n-1 A(j,k).$$
answered Jul 29 at 7:55
Robert Z
83.7k954122
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In the two given double sums, we are summing over the same set of couples $$S_n:=(j,k)inmathbbNtimes mathbbN: 0leq kleq jleq n-1.$$
Hence for ANY double sequence $(A(j,k))_j,k$,
$$sum_j=0^n-1sum_k=0^j A(j,k)=sum_(j,k)in S_nA(j,k)=
sum_k=0^n-1sum_j=k^n-1 A(j,k).$$
answered Jul 29 at 7:55
Robert Z
83.7k954122
add a comment |Â
up vote
2
down vote
accepted
In the two given double sums, we are summing over the same set of couples $$S_n:=(j,k)inmathbbNtimes mathbbN: 0leq kleq jleq n-1.$$
Hence for ANY double sequence $(A(j,k))_j,k$,
$$sum_j=0^n-1sum_k=0^j A(j,k)=sum_(j,k)in S_nA(j,k)=
sum_k=0^n-1sum_j=k^n-1 A(j,k).$$
answered Jul 29 at 7:55
Robert Z
83.7k954122
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In the two given double sums, we are summing over the same set of couples $$S_n:=(j,k)inmathbbNtimes mathbbN: 0leq kleq jleq n-1.$$
Hence for ANY double sequence $(A(j,k))_j,k$,
$$sum_j=0^n-1sum_k=0^j A(j,k)=sum_(j,k)in S_nA(j,k)=
sum_k=0^n-1sum_j=k^n-1 A(j,k).$$
answered Jul 29 at 7:55
Robert Z
83.7k954122
In the two given double sums, we are summing over the same set of couples $$S_n:=(j,k)inmathbbNtimes mathbbN: 0leq kleq jleq n-1.$$
Hence for ANY double sequence $(A(j,k))_j,k$,
$$sum_j=0^n-1sum_k=0^j A(j,k)=sum_(j,k)in S_nA(j,k)=
sum_k=0^n-1sum_j=k^n-1 A(j,k).$$
answered Jul 29 at 7:55
Robert Z
83.7k954122
edited Jul 29 at 8:03
answered Jul 29 at 7:55
Robert Z
83.7k954122
answered Jul 29 at 7:55
Robert Z
83.7k954122
answered Jul 29 at 7:55
Robert Z
83.7k954122
83.7k954122
add a comment |Â
add a comment |Â
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1
You are summing over the set of couples $(k,j): 0leq kleq jleq n-1$.
– Robert Z
Jul 29 at 7:48
How does this prove the above statement?
– Ole Petersen
Jul 29 at 7:52