I need help simplifying a sum problem that involves a binomial raised to a power
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I have come across a problem in my homework that describes the sum of a binomial squared, and I can't think of a way to simplify it. I have an idea that it would involve $fracleft(nright)left(n+1right)left(2n+1right)6$, but I can't seem to format the equation in a way that would be helpful. Any help would be much appreciated. There are two problems below, a solution to either would be amazing.
$sum_i=0^n-1fracleft(fracleft(n+1right)2-iright)^2n$
$sum_i=0^n-1left(fracleft(n+1right)2-iright)^2$
summation binomial-coefficients exponential-sum
asked Aug 19 at 5:16
Ryan Shesler
33
add a comment |Â
up vote
-2
down vote
favorite
I have come across a problem in my homework that describes the sum of a binomial squared, and I can't think of a way to simplify it. I have an idea that it would involve $fracleft(nright)left(n+1right)left(2n+1right)6$, but I can't seem to format the equation in a way that would be helpful. Any help would be much appreciated. There are two problems below, a solution to either would be amazing.
$sum_i=0^n-1fracleft(fracleft(n+1right)2-iright)^2n$
$sum_i=0^n-1left(fracleft(n+1right)2-iright)^2$
summation binomial-coefficients exponential-sum
asked Aug 19 at 5:16
Ryan Shesler
33
Square the binomial first: $displaystyle;sum_i=0^n-1left(fracleft(n+1right)^24-(n+1),i+i^2right) = ldots;$
– dxiv
Aug 19 at 5:27
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I have come across a problem in my homework that describes the sum of a binomial squared, and I can't think of a way to simplify it. I have an idea that it would involve $fracleft(nright)left(n+1right)left(2n+1right)6$, but I can't seem to format the equation in a way that would be helpful. Any help would be much appreciated. There are two problems below, a solution to either would be amazing.
$sum_i=0^n-1fracleft(fracleft(n+1right)2-iright)^2n$
$sum_i=0^n-1left(fracleft(n+1right)2-iright)^2$
summation binomial-coefficients exponential-sum
asked Aug 19 at 5:16
Ryan Shesler
33
I have come across a problem in my homework that describes the sum of a binomial squared, and I can't think of a way to simplify it. I have an idea that it would involve $fracleft(nright)left(n+1right)left(2n+1right)6$, but I can't seem to format the equation in a way that would be helpful. Any help would be much appreciated. There are two problems below, a solution to either would be amazing.
$sum_i=0^n-1fracleft(fracleft(n+1right)2-iright)^2n$
$sum_i=0^n-1left(fracleft(n+1right)2-iright)^2$
summation binomial-coefficients exponential-sum
asked Aug 19 at 5:16
Ryan Shesler
33
edited Aug 19 at 5:32
asked Aug 19 at 5:16
Ryan Shesler
33
asked Aug 19 at 5:16
Ryan Shesler
33
asked Aug 19 at 5:16
Ryan Shesler
33
33
Square the binomial first: $displaystyle;sum_i=0^n-1left(fracleft(n+1right)^24-(n+1),i+i^2right) = ldots;$
– dxiv
Aug 19 at 5:27
add a comment |Â
Square the binomial first: $displaystyle;sum_i=0^n-1left(fracleft(n+1right)^24-(n+1),i+i^2right) = ldots;$
– dxiv
Aug 19 at 5:27
Square the binomial first: $displaystyle;sum_i=0^n-1left(fracleft(n+1right)^24-(n+1),i+i^2right) = ldots;$
– dxiv
Aug 19 at 5:27
Square the binomial first: $displaystyle;sum_i=0^n-1left(fracleft(n+1right)^24-(n+1),i+i^2right) = ldots;$
– dxiv
Aug 19 at 5:27
add a comment |Â
1 Answer
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$beginarray\
s(n)
&=sum_i=0^n-1left(fracleft(n+1right)2-iright)^2\
&=frac14sum_i=0^n-1left(n+1-2iright)^2\
&=frac14sum_i=0^n-1((n+1)^2-4(n+1)i+4i^2)\
&=frac14sum_i=0^n-1(n+1)^2-sum_i=0^n-1(n+1)i+sum_i=0^n-1i^2\
&=frac14n(n+1)^2-frac12 n(n-1)(n+1)+frac16 (n-1)n(2n-1)\
&=dfracn12left( 3(n+1)^2-6(n-1)(n+1)+2 (n-1)(2n-1)right)\
&=dfracn12left( 3n^2+6n+3-6(n^2-1)+2(2n^2-3n+1)right)\
&=dfracn12left( 3n^2+6n+3-6n^2+6+4n^2-6n+2right)\
&=dfracn12left( n^2+11right)\
endarray
$
answered Aug 19 at 5:51
marty cohen
69.6k446122
Hi! Thank you for the help! The problem checks, but I was just curious as to how you solved each of the sums in step 4.
– Ryan Shesler
Aug 19 at 16:33
does it have something to do with sum of a series, and if so how? @martycohen
– Ryan Shesler
Aug 19 at 16:46
Well known sums: $sum_i=0^n-1 i = n(n-1)/2, sum_i=0^n-1 i^2 = n(n-1)(2n-1)/6$.
– marty cohen
Aug 19 at 22:33
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$beginarray\
s(n)
&=sum_i=0^n-1left(fracleft(n+1right)2-iright)^2\
&=frac14sum_i=0^n-1left(n+1-2iright)^2\
&=frac14sum_i=0^n-1((n+1)^2-4(n+1)i+4i^2)\
&=frac14sum_i=0^n-1(n+1)^2-sum_i=0^n-1(n+1)i+sum_i=0^n-1i^2\
&=frac14n(n+1)^2-frac12 n(n-1)(n+1)+frac16 (n-1)n(2n-1)\
&=dfracn12left( 3(n+1)^2-6(n-1)(n+1)+2 (n-1)(2n-1)right)\
&=dfracn12left( 3n^2+6n+3-6(n^2-1)+2(2n^2-3n+1)right)\
&=dfracn12left( 3n^2+6n+3-6n^2+6+4n^2-6n+2right)\
&=dfracn12left( n^2+11right)\
endarray
$
answered Aug 19 at 5:51
marty cohen
69.6k446122
Hi! Thank you for the help! The problem checks, but I was just curious as to how you solved each of the sums in step 4.
– Ryan Shesler
Aug 19 at 16:33
does it have something to do with sum of a series, and if so how? @martycohen
– Ryan Shesler
Aug 19 at 16:46
Well known sums: $sum_i=0^n-1 i = n(n-1)/2, sum_i=0^n-1 i^2 = n(n-1)(2n-1)/6$.
– marty cohen
Aug 19 at 22:33
add a comment |Â
up vote
0
down vote
accepted
$beginarray\
s(n)
&=sum_i=0^n-1left(fracleft(n+1right)2-iright)^2\
&=frac14sum_i=0^n-1left(n+1-2iright)^2\
&=frac14sum_i=0^n-1((n+1)^2-4(n+1)i+4i^2)\
&=frac14sum_i=0^n-1(n+1)^2-sum_i=0^n-1(n+1)i+sum_i=0^n-1i^2\
&=frac14n(n+1)^2-frac12 n(n-1)(n+1)+frac16 (n-1)n(2n-1)\
&=dfracn12left( 3(n+1)^2-6(n-1)(n+1)+2 (n-1)(2n-1)right)\
&=dfracn12left( 3n^2+6n+3-6(n^2-1)+2(2n^2-3n+1)right)\
&=dfracn12left( 3n^2+6n+3-6n^2+6+4n^2-6n+2right)\
&=dfracn12left( n^2+11right)\
endarray
$
answered Aug 19 at 5:51
marty cohen
69.6k446122
Hi! Thank you for the help! The problem checks, but I was just curious as to how you solved each of the sums in step 4.
– Ryan Shesler
Aug 19 at 16:33
does it have something to do with sum of a series, and if so how? @martycohen
– Ryan Shesler
Aug 19 at 16:46
Well known sums: $sum_i=0^n-1 i = n(n-1)/2, sum_i=0^n-1 i^2 = n(n-1)(2n-1)/6$.
– marty cohen
Aug 19 at 22:33
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$beginarray\
s(n)
&=sum_i=0^n-1left(fracleft(n+1right)2-iright)^2\
&=frac14sum_i=0^n-1left(n+1-2iright)^2\
&=frac14sum_i=0^n-1((n+1)^2-4(n+1)i+4i^2)\
&=frac14sum_i=0^n-1(n+1)^2-sum_i=0^n-1(n+1)i+sum_i=0^n-1i^2\
&=frac14n(n+1)^2-frac12 n(n-1)(n+1)+frac16 (n-1)n(2n-1)\
&=dfracn12left( 3(n+1)^2-6(n-1)(n+1)+2 (n-1)(2n-1)right)\
&=dfracn12left( 3n^2+6n+3-6(n^2-1)+2(2n^2-3n+1)right)\
&=dfracn12left( 3n^2+6n+3-6n^2+6+4n^2-6n+2right)\
&=dfracn12left( n^2+11right)\
endarray
$
answered Aug 19 at 5:51
marty cohen
69.6k446122
$beginarray\
s(n)
&=sum_i=0^n-1left(fracleft(n+1right)2-iright)^2\
&=frac14sum_i=0^n-1left(n+1-2iright)^2\
&=frac14sum_i=0^n-1((n+1)^2-4(n+1)i+4i^2)\
&=frac14sum_i=0^n-1(n+1)^2-sum_i=0^n-1(n+1)i+sum_i=0^n-1i^2\
&=frac14n(n+1)^2-frac12 n(n-1)(n+1)+frac16 (n-1)n(2n-1)\
&=dfracn12left( 3(n+1)^2-6(n-1)(n+1)+2 (n-1)(2n-1)right)\
&=dfracn12left( 3n^2+6n+3-6(n^2-1)+2(2n^2-3n+1)right)\
&=dfracn12left( 3n^2+6n+3-6n^2+6+4n^2-6n+2right)\
&=dfracn12left( n^2+11right)\
endarray
$
answered Aug 19 at 5:51
marty cohen
69.6k446122
answered Aug 19 at 5:51
marty cohen
69.6k446122
answered Aug 19 at 5:51
marty cohen
69.6k446122
answered Aug 19 at 5:51
marty cohen
69.6k446122
69.6k446122
Hi! Thank you for the help! The problem checks, but I was just curious as to how you solved each of the sums in step 4.
– Ryan Shesler
Aug 19 at 16:33
does it have something to do with sum of a series, and if so how? @martycohen
– Ryan Shesler
Aug 19 at 16:46
Well known sums: $sum_i=0^n-1 i = n(n-1)/2, sum_i=0^n-1 i^2 = n(n-1)(2n-1)/6$.
– marty cohen
Aug 19 at 22:33
add a comment |Â
Hi! Thank you for the help! The problem checks, but I was just curious as to how you solved each of the sums in step 4.
– Ryan Shesler
Aug 19 at 16:33
does it have something to do with sum of a series, and if so how? @martycohen
– Ryan Shesler
Aug 19 at 16:46
Well known sums: $sum_i=0^n-1 i = n(n-1)/2, sum_i=0^n-1 i^2 = n(n-1)(2n-1)/6$.
– marty cohen
Aug 19 at 22:33
Hi! Thank you for the help! The problem checks, but I was just curious as to how you solved each of the sums in step 4.
– Ryan Shesler
Aug 19 at 16:33
Hi! Thank you for the help! The problem checks, but I was just curious as to how you solved each of the sums in step 4.
– Ryan Shesler
Aug 19 at 16:33
does it have something to do with sum of a series, and if so how? @martycohen
– Ryan Shesler
Aug 19 at 16:46
does it have something to do with sum of a series, and if so how? @martycohen
– Ryan Shesler
Aug 19 at 16:46
Well known sums: $sum_i=0^n-1 i = n(n-1)/2, sum_i=0^n-1 i^2 = n(n-1)(2n-1)/6$.
– marty cohen
Aug 19 at 22:33
Well known sums: $sum_i=0^n-1 i = n(n-1)/2, sum_i=0^n-1 i^2 = n(n-1)(2n-1)/6$.
– marty cohen
Aug 19 at 22:33
add a comment |Â
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Square the binomial first: $displaystyle;sum_i=0^n-1left(fracleft(n+1right)^24-(n+1),i+i^2right) = ldots;$
– dxiv
Aug 19 at 5:27