Mixing
Find the sum of this series $sum_{n=1}^{infty}frac{2n^{2}-3n+4}{2^{n}}$
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Can someone provide help in finding the sum of this series? $$sum_{n=1}^{infty}frac{2n^{2}-3n+4}{2^{n}}$$
real-analysis sequences-and-series
edited yesterday
Lorenzo B.
1,6222419
asked yesterday
dimpap
657
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up vote
0
down vote
favorite
Can someone provide help in finding the sum of this series? $$sum_{n=1}^{infty}frac{2n^{2}-3n+4}{2^{n}}$$
real-analysis sequences-and-series
edited yesterday
Lorenzo B.
1,6222419
asked yesterday
dimpap
657
By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
– Jack D'Aurizio
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can someone provide help in finding the sum of this series? $$sum_{n=1}^{infty}frac{2n^{2}-3n+4}{2^{n}}$$
real-analysis sequences-and-series
edited yesterday
Lorenzo B.
1,6222419
asked yesterday
dimpap
657
Can someone provide help in finding the sum of this series? $$sum_{n=1}^{infty}frac{2n^{2}-3n+4}{2^{n}}$$
real-analysis sequences-and-series
real-analysis sequences-and-series
edited yesterday
Lorenzo B.
1,6222419
asked yesterday
dimpap
657
edited yesterday
Lorenzo B.
1,6222419
asked yesterday
dimpap
657
edited yesterday
Lorenzo B.
1,6222419
edited yesterday
Lorenzo B.
1,6222419
edited yesterday
Lorenzo B.
1,6222419
1,6222419
asked yesterday
dimpap
657
asked yesterday
dimpap
657
asked yesterday
dimpap
657
657
By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
– Jack D'Aurizio
yesterday
add a comment |
By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
– Jack D'Aurizio
yesterday
By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
– Jack D'Aurizio
yesterday
By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
– Jack D'Aurizio
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Let $f(m)$
$$=dfrac{a+bm+cm^2}{2^m}$$
Set $dfrac{2n^2-3n+4}{2^n}=f(n+1)-f(n)$ and compare the coefficients of $n,n^2$ and the constants to find $a,b,c$
Use https://en.m.wikipedia.org/wiki/Telescoping_series
answered yesterday
lab bhattacharjee
220k15154270
Super answer. thanks.
– hamam_Abdallah
yesterday
See also math.stackexchange.com/questions/593996/…
– lab bhattacharjee
yesterday
thanks very much!
– dimpap
yesterday
add a comment |
up vote
0
down vote
hint
For $-2<x<2$,
$$sum_{n=0}^{+infty}frac{x^n}{2^n}=frac{2}{2-x}$$
Think differentiating both sides and make $x=1$.
Observe that the numerator can be written as
$$2n(n-1)-n+4$$
answered yesterday
hamam_Abdallah
36.5k21533
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $f(m)$
$$=dfrac{a+bm+cm^2}{2^m}$$
Set $dfrac{2n^2-3n+4}{2^n}=f(n+1)-f(n)$ and compare the coefficients of $n,n^2$ and the constants to find $a,b,c$
Use https://en.m.wikipedia.org/wiki/Telescoping_series
answered yesterday
lab bhattacharjee
220k15154270
Super answer. thanks.
– hamam_Abdallah
yesterday
See also math.stackexchange.com/questions/593996/…
– lab bhattacharjee
yesterday
thanks very much!
– dimpap
yesterday
add a comment |
up vote
2
down vote
accepted
Let $f(m)$
$$=dfrac{a+bm+cm^2}{2^m}$$
Set $dfrac{2n^2-3n+4}{2^n}=f(n+1)-f(n)$ and compare the coefficients of $n,n^2$ and the constants to find $a,b,c$
Use https://en.m.wikipedia.org/wiki/Telescoping_series
answered yesterday
lab bhattacharjee
220k15154270
Super answer. thanks.
– hamam_Abdallah
yesterday
See also math.stackexchange.com/questions/593996/…
– lab bhattacharjee
yesterday
thanks very much!
– dimpap
yesterday
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $f(m)$
$$=dfrac{a+bm+cm^2}{2^m}$$
Set $dfrac{2n^2-3n+4}{2^n}=f(n+1)-f(n)$ and compare the coefficients of $n,n^2$ and the constants to find $a,b,c$
Use https://en.m.wikipedia.org/wiki/Telescoping_series
answered yesterday
lab bhattacharjee
220k15154270
Let $f(m)$
$$=dfrac{a+bm+cm^2}{2^m}$$
Set $dfrac{2n^2-3n+4}{2^n}=f(n+1)-f(n)$ and compare the coefficients of $n,n^2$ and the constants to find $a,b,c$
Use https://en.m.wikipedia.org/wiki/Telescoping_series
answered yesterday
lab bhattacharjee
220k15154270
answered yesterday
lab bhattacharjee
220k15154270
answered yesterday
lab bhattacharjee
220k15154270
answered yesterday
lab bhattacharjee
220k15154270
220k15154270
Super answer. thanks.
– hamam_Abdallah
yesterday
See also math.stackexchange.com/questions/593996/…
– lab bhattacharjee
yesterday
thanks very much!
– dimpap
yesterday
add a comment |
Super answer. thanks.
– hamam_Abdallah
yesterday
See also math.stackexchange.com/questions/593996/…
– lab bhattacharjee
yesterday
thanks very much!
– dimpap
yesterday
Super answer. thanks.
– hamam_Abdallah
yesterday
Super answer. thanks.
– hamam_Abdallah
yesterday
See also math.stackexchange.com/questions/593996/…
– lab bhattacharjee
yesterday
See also math.stackexchange.com/questions/593996/…
– lab bhattacharjee
yesterday
thanks very much!
– dimpap
yesterday
thanks very much!
– dimpap
yesterday
add a comment |
up vote
0
down vote
hint
For $-2<x<2$,
$$sum_{n=0}^{+infty}frac{x^n}{2^n}=frac{2}{2-x}$$
Think differentiating both sides and make $x=1$.
Observe that the numerator can be written as
$$2n(n-1)-n+4$$
answered yesterday
hamam_Abdallah
36.5k21533
add a comment |
up vote
0
down vote
hint
For $-2<x<2$,
$$sum_{n=0}^{+infty}frac{x^n}{2^n}=frac{2}{2-x}$$
Think differentiating both sides and make $x=1$.
Observe that the numerator can be written as
$$2n(n-1)-n+4$$
answered yesterday
hamam_Abdallah
36.5k21533
add a comment |
up vote
0
down vote
up vote
0
down vote
hint
For $-2<x<2$,
$$sum_{n=0}^{+infty}frac{x^n}{2^n}=frac{2}{2-x}$$
Think differentiating both sides and make $x=1$.
Observe that the numerator can be written as
$$2n(n-1)-n+4$$
answered yesterday
hamam_Abdallah
36.5k21533
hint
For $-2<x<2$,
$$sum_{n=0}^{+infty}frac{x^n}{2^n}=frac{2}{2-x}$$
Think differentiating both sides and make $x=1$.
Observe that the numerator can be written as
$$2n(n-1)-n+4$$
answered yesterday
hamam_Abdallah
36.5k21533
edited yesterday
answered yesterday
hamam_Abdallah
36.5k21533
answered yesterday
hamam_Abdallah
36.5k21533
answered yesterday
hamam_Abdallah
36.5k21533
36.5k21533
add a comment |
add a comment |
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By stars and bars we have $$ frac{1}{(1-x)^{n+1}}=sum_{mgeq 0}binom{m+n}{n}x^m $$ for any $xin(-1,1)$ and any $ninmathbb{N}$. By reindexing $$ sum_{ngeq 1}frac{2n^2-3n+4}{2^n} = sum_{mgeq 0}frac{2m^2+m+3}{2^{m+1}}=frac{1}{2}sum_{mgeq 0}frac{4binom{m+2}{2}-5binom{m+1}{1}+4binom{m+0}{0}}{2^m} $$ hence the LHS equals $$ frac{1}{2}left[4cdot 2^3-5cdot 2^2+4cdot 2^1right]=color{red}{10}. $$
– Jack D'Aurizio
yesterday