Showing a sum is positiveFinding Binomial expansion of a radicalSimplify the Expression $sum _ k=0 ^ n binomnki^k3^k-n $Alternative combinatorial proof for $sumlimits_r=0^nbinomnrbinomm+rn=sumlimits_r=0^nbinomnrbinommr2^r$Proof by induction, binomial coefficientApproximating a binomial sum over a simplexHow to expand $sqrtx^6+1$ using Maclaurin's seriesSum of $m choose j$ multiplied by $2^2^j$How to show that $sumlimits_k=0^n (-1)^ktfrac nchoosek x+kchoosek = fracxx+n$Finite sum with inverse binomialShowing an alternating sum is positive
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Showing a sum is positive
Finding Binomial expansion of a radicalSimplify the Expression $sum _ k=0 ^ n binomnki^k3^k-n $Alternative combinatorial proof for $sumlimits_r=0^nbinomnrbinomm+rn=sumlimits_r=0^nbinomnrbinommr2^r$Proof by induction, binomial coefficientApproximating a binomial sum over a simplexHow to expand $sqrtx^6+1$ using Maclaurin's seriesSum of $m choose j$ multiplied by $2^2^j$How to show that $sumlimits_k=0^n (-1)^ktfrac nchoosek x+kchoosek = fracxx+n$Finite sum with inverse binomialShowing an alternating sum is positive
$begingroup$
Show that the sum$$sum_k=0^n n choose kfrac(-1)^kn+k+1$$ is a positive rational number.
It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.
combinatorics summation binomial-coefficients binomial-ideals
asked 1 hour ago
Hitendra KumarHitendra Kumar
606
$endgroup$
add a comment |
$begingroup$
Show that the sum$$sum_k=0^n n choose kfrac(-1)^kn+k+1$$ is a positive rational number.
It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.
combinatorics summation binomial-coefficients binomial-ideals
asked 1 hour ago
Hitendra KumarHitendra Kumar
606
$endgroup$
1
$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago
add a comment |
$begingroup$
Show that the sum$$sum_k=0^n n choose kfrac(-1)^kn+k+1$$ is a positive rational number.
It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.
combinatorics summation binomial-coefficients binomial-ideals
asked 1 hour ago
Hitendra KumarHitendra Kumar
606
$endgroup$
Show that the sum$$sum_k=0^n n choose kfrac(-1)^kn+k+1$$ is a positive rational number.
It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.
combinatorics summation binomial-coefficients binomial-ideals
combinatorics summation binomial-coefficients binomial-ideals
asked 1 hour ago
Hitendra KumarHitendra Kumar
606
asked 1 hour ago
Hitendra KumarHitendra Kumar
606
edited 1 hour ago
Hitendra Kumar
asked 1 hour ago
Hitendra KumarHitendra Kumar
606
asked 1 hour ago
Hitendra KumarHitendra Kumar
606
asked 1 hour ago
Hitendra KumarHitendra Kumar
606
606
1
$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago
add a comment |
1
$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago
1
1
$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago
$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Direct proof:
$$beginsplit
sum_k=0^n nchoose kfrac(-1)^kn+k+1 &=sum_k=0^n nchoose k(-1)^kint_0^1 x^n+kdx\
&=int_0^1x^nsum_k=0^n nchoose k(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
endsplit$$
The latter is clearly a positive number.
answered 1 hour ago
Stefan LafonStefan Lafon
3,00019
$endgroup$
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
35 mins ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
34 mins ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
34 mins ago
1
$begingroup$
It's a "known trick" that $frac 1 p+1 = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
28 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
18 mins ago
add a comment |
$begingroup$
When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.
When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.
.....
Get it?
answered 1 hour ago
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Direct proof:
$$beginsplit
sum_k=0^n nchoose kfrac(-1)^kn+k+1 &=sum_k=0^n nchoose k(-1)^kint_0^1 x^n+kdx\
&=int_0^1x^nsum_k=0^n nchoose k(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
endsplit$$
The latter is clearly a positive number.
answered 1 hour ago
Stefan LafonStefan Lafon
3,00019
$endgroup$
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
35 mins ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
34 mins ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
34 mins ago
1
$begingroup$
It's a "known trick" that $frac 1 p+1 = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
28 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
18 mins ago
add a comment |
$begingroup$
Direct proof:
$$beginsplit
sum_k=0^n nchoose kfrac(-1)^kn+k+1 &=sum_k=0^n nchoose k(-1)^kint_0^1 x^n+kdx\
&=int_0^1x^nsum_k=0^n nchoose k(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
endsplit$$
The latter is clearly a positive number.
answered 1 hour ago
Stefan LafonStefan Lafon
3,00019
$endgroup$
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
35 mins ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
34 mins ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
34 mins ago
1
$begingroup$
It's a "known trick" that $frac 1 p+1 = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
28 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
18 mins ago
add a comment |
$begingroup$
Direct proof:
$$beginsplit
sum_k=0^n nchoose kfrac(-1)^kn+k+1 &=sum_k=0^n nchoose k(-1)^kint_0^1 x^n+kdx\
&=int_0^1x^nsum_k=0^n nchoose k(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
endsplit$$
The latter is clearly a positive number.
answered 1 hour ago
Stefan LafonStefan Lafon
3,00019
$endgroup$
Direct proof:
$$beginsplit
sum_k=0^n nchoose kfrac(-1)^kn+k+1 &=sum_k=0^n nchoose k(-1)^kint_0^1 x^n+kdx\
&=int_0^1x^nsum_k=0^n nchoose k(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
endsplit$$
The latter is clearly a positive number.
answered 1 hour ago
Stefan LafonStefan Lafon
3,00019
answered 1 hour ago
Stefan LafonStefan Lafon
3,00019
answered 1 hour ago
Stefan LafonStefan Lafon
3,00019
answered 1 hour ago
Stefan LafonStefan Lafon
3,00019
3,00019
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
35 mins ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
34 mins ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
34 mins ago
1
$begingroup$
It's a "known trick" that $frac 1 p+1 = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
28 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
18 mins ago
add a comment |
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
35 mins ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
34 mins ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
34 mins ago
1
$begingroup$
It's a "known trick" that $frac 1 p+1 = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
28 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
18 mins ago
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
35 mins ago
$begingroup$
Thanks,I got it.
$endgroup$
– Hitendra Kumar
35 mins ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
34 mins ago
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
34 mins ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
34 mins ago
$begingroup$
How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
$endgroup$
– NoChance
34 mins ago
1
1
$begingroup$
It's a "known trick" that $frac 1 p+1 = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
28 mins ago
$begingroup$
It's a "known trick" that $frac 1 p+1 = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
$endgroup$
– Stefan Lafon
28 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
18 mins ago
$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
18 mins ago
add a comment |
$begingroup$
When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.
When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.
.....
Get it?
answered 1 hour ago
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
add a comment |
$begingroup$
When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.
When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.
.....
Get it?
answered 1 hour ago
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
add a comment |
$begingroup$
When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.
When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.
.....
Get it?
answered 1 hour ago
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.
When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.
.....
Get it?
answered 1 hour ago
David G. StorkDavid G. Stork
11.1k41432
answered 1 hour ago
David G. StorkDavid G. Stork
11.1k41432
answered 1 hour ago
David G. StorkDavid G. Stork
11.1k41432
answered 1 hour ago
David G. StorkDavid G. Stork
11.1k41432
11.1k41432
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
add a comment |
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
$begingroup$
sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
$endgroup$
– Hitendra Kumar
1 hour ago
add a comment |
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$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
1 hour ago