Calculate $sum_{k=1}^{2007} frac{5^{2008}}{25^k+5^{2008}}$
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$sum_{k=1}^{2007} frac{5^{2008}}{25^k+5^{2008}}$. I know that solution is $frac{2007}{2}$ and that I can use a Gaussian method.
If I use Gaussian method I do not get some sum that help me. if I put $n=2008$, then $Sn=sum_{k=1}^n frac{5^{n+1}}{25^k+5^{n+1}}$. and $Sn=sum_{k=1}^n frac{5^{n}}{25^{k-n}+5^{n+1}}$. So then $2Sn=sum_{k=1}^n frac{5^{n+1}}{25^k+5^{n+1}}+frac{5^{n+1}}{25^{n-k}+5^{n+1}}$. do you have something better?
discrete-mathematics summation
asked Nov 21 at 17:46
Marko Škorić
67910
add a comment |
up vote
1
down vote
favorite
$sum_{k=1}^{2007} frac{5^{2008}}{25^k+5^{2008}}$. I know that solution is $frac{2007}{2}$ and that I can use a Gaussian method.
If I use Gaussian method I do not get some sum that help me. if I put $n=2008$, then $Sn=sum_{k=1}^n frac{5^{n+1}}{25^k+5^{n+1}}$. and $Sn=sum_{k=1}^n frac{5^{n}}{25^{k-n}+5^{n+1}}$. So then $2Sn=sum_{k=1}^n frac{5^{n+1}}{25^k+5^{n+1}}+frac{5^{n+1}}{25^{n-k}+5^{n+1}}$. do you have something better?
discrete-mathematics summation
asked Nov 21 at 17:46
Marko Škorić
67910
What is the "Gaussian method"?
– user587192
Nov 21 at 17:52
gaus method for calculate for sum but i do not know is this good translation
– Marko Škorić
Nov 21 at 17:57
3
math.stackexchange.com/questions/2541131/…
– lab bhattacharjee
Nov 21 at 18:06
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$sum_{k=1}^{2007} frac{5^{2008}}{25^k+5^{2008}}$. I know that solution is $frac{2007}{2}$ and that I can use a Gaussian method.
If I use Gaussian method I do not get some sum that help me. if I put $n=2008$, then $Sn=sum_{k=1}^n frac{5^{n+1}}{25^k+5^{n+1}}$. and $Sn=sum_{k=1}^n frac{5^{n}}{25^{k-n}+5^{n+1}}$. So then $2Sn=sum_{k=1}^n frac{5^{n+1}}{25^k+5^{n+1}}+frac{5^{n+1}}{25^{n-k}+5^{n+1}}$. do you have something better?
discrete-mathematics summation
asked Nov 21 at 17:46
Marko Škorić
67910
$sum_{k=1}^{2007} frac{5^{2008}}{25^k+5^{2008}}$. I know that solution is $frac{2007}{2}$ and that I can use a Gaussian method.
If I use Gaussian method I do not get some sum that help me. if I put $n=2008$, then $Sn=sum_{k=1}^n frac{5^{n+1}}{25^k+5^{n+1}}$. and $Sn=sum_{k=1}^n frac{5^{n}}{25^{k-n}+5^{n+1}}$. So then $2Sn=sum_{k=1}^n frac{5^{n+1}}{25^k+5^{n+1}}+frac{5^{n+1}}{25^{n-k}+5^{n+1}}$. do you have something better?
discrete-mathematics summation
discrete-mathematics summation
asked Nov 21 at 17:46
Marko Škorić
67910
asked Nov 21 at 17:46
Marko Škorić
67910
edited Nov 21 at 17:57
asked Nov 21 at 17:46
Marko Škorić
67910
asked Nov 21 at 17:46
Marko Škorić
67910
asked Nov 21 at 17:46
Marko Škorić
67910
67910
What is the "Gaussian method"?
– user587192
Nov 21 at 17:52
gaus method for calculate for sum but i do not know is this good translation
– Marko Škorić
Nov 21 at 17:57
3
math.stackexchange.com/questions/2541131/…
– lab bhattacharjee
Nov 21 at 18:06
add a comment |
What is the "Gaussian method"?
– user587192
Nov 21 at 17:52
gaus method for calculate for sum but i do not know is this good translation
– Marko Škorić
Nov 21 at 17:57
3
math.stackexchange.com/questions/2541131/…
– lab bhattacharjee
Nov 21 at 18:06
What is the "Gaussian method"?
– user587192
Nov 21 at 17:52
What is the "Gaussian method"?
– user587192
Nov 21 at 17:52
gaus method for calculate for sum but i do not know is this good translation
– Marko Škorić
Nov 21 at 17:57
gaus method for calculate for sum but i do not know is this good translation
– Marko Škorić
Nov 21 at 17:57
3
3
math.stackexchange.com/questions/2541131/…
– lab bhattacharjee
Nov 21 at 18:06
math.stackexchange.com/questions/2541131/…
– lab bhattacharjee
Nov 21 at 18:06
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Using $frac{1}{1+q}+frac{1}{1+1/q}=1$, $$sum_{k=1}^nfrac{1}{1+5^{2k-n-1}}=frac{1}{2}sum_{k=1}^nbigg(frac{1}{1+5^{2k-n-1}}+frac{1}{1+5^{2(n+1-k)-n-1}}bigg)\=frac{1}{2}sum_{k=1}^nbigg(frac{1}{1+5^{2k-n-1}}+frac{1}{1+5^{n+1-2k}}bigg)=frac{1}{2}sum_{k=1}^n 1=frac{n}{2}.$$
answered Nov 21 at 18:10
J.G.
18.7k21932
add a comment |
up vote
0
down vote
Write $$sum_{k=1}^{2007} frac{5^{2008}}{25^k+5^{2008}}=sum_{k=1}^{2007}{25^{1004}over 25^k+25^{1004}} = sum_{k=1}^{2007}{1over 25^{k-1004}+1} =sum_{j=-1003}^{1003}{1over 2^{-j}+1}$$
Now try your pairing strategy.
answered Nov 21 at 18:11
saulspatz
12.9k21327
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Using $frac{1}{1+q}+frac{1}{1+1/q}=1$, $$sum_{k=1}^nfrac{1}{1+5^{2k-n-1}}=frac{1}{2}sum_{k=1}^nbigg(frac{1}{1+5^{2k-n-1}}+frac{1}{1+5^{2(n+1-k)-n-1}}bigg)\=frac{1}{2}sum_{k=1}^nbigg(frac{1}{1+5^{2k-n-1}}+frac{1}{1+5^{n+1-2k}}bigg)=frac{1}{2}sum_{k=1}^n 1=frac{n}{2}.$$
answered Nov 21 at 18:10
J.G.
18.7k21932
add a comment |
up vote
5
down vote
accepted
Using $frac{1}{1+q}+frac{1}{1+1/q}=1$, $$sum_{k=1}^nfrac{1}{1+5^{2k-n-1}}=frac{1}{2}sum_{k=1}^nbigg(frac{1}{1+5^{2k-n-1}}+frac{1}{1+5^{2(n+1-k)-n-1}}bigg)\=frac{1}{2}sum_{k=1}^nbigg(frac{1}{1+5^{2k-n-1}}+frac{1}{1+5^{n+1-2k}}bigg)=frac{1}{2}sum_{k=1}^n 1=frac{n}{2}.$$
answered Nov 21 at 18:10
J.G.
18.7k21932
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Using $frac{1}{1+q}+frac{1}{1+1/q}=1$, $$sum_{k=1}^nfrac{1}{1+5^{2k-n-1}}=frac{1}{2}sum_{k=1}^nbigg(frac{1}{1+5^{2k-n-1}}+frac{1}{1+5^{2(n+1-k)-n-1}}bigg)\=frac{1}{2}sum_{k=1}^nbigg(frac{1}{1+5^{2k-n-1}}+frac{1}{1+5^{n+1-2k}}bigg)=frac{1}{2}sum_{k=1}^n 1=frac{n}{2}.$$
answered Nov 21 at 18:10
J.G.
18.7k21932
Using $frac{1}{1+q}+frac{1}{1+1/q}=1$, $$sum_{k=1}^nfrac{1}{1+5^{2k-n-1}}=frac{1}{2}sum_{k=1}^nbigg(frac{1}{1+5^{2k-n-1}}+frac{1}{1+5^{2(n+1-k)-n-1}}bigg)\=frac{1}{2}sum_{k=1}^nbigg(frac{1}{1+5^{2k-n-1}}+frac{1}{1+5^{n+1-2k}}bigg)=frac{1}{2}sum_{k=1}^n 1=frac{n}{2}.$$
answered Nov 21 at 18:10
J.G.
18.7k21932
answered Nov 21 at 18:10
J.G.
18.7k21932
answered Nov 21 at 18:10
J.G.
18.7k21932
answered Nov 21 at 18:10
J.G.
18.7k21932
18.7k21932
add a comment |
add a comment |
up vote
0
down vote
Write $$sum_{k=1}^{2007} frac{5^{2008}}{25^k+5^{2008}}=sum_{k=1}^{2007}{25^{1004}over 25^k+25^{1004}} = sum_{k=1}^{2007}{1over 25^{k-1004}+1} =sum_{j=-1003}^{1003}{1over 2^{-j}+1}$$
Now try your pairing strategy.
answered Nov 21 at 18:11
saulspatz
12.9k21327
add a comment |
up vote
0
down vote
Write $$sum_{k=1}^{2007} frac{5^{2008}}{25^k+5^{2008}}=sum_{k=1}^{2007}{25^{1004}over 25^k+25^{1004}} = sum_{k=1}^{2007}{1over 25^{k-1004}+1} =sum_{j=-1003}^{1003}{1over 2^{-j}+1}$$
Now try your pairing strategy.
answered Nov 21 at 18:11
saulspatz
12.9k21327
add a comment |
up vote
0
down vote
up vote
0
down vote
Write $$sum_{k=1}^{2007} frac{5^{2008}}{25^k+5^{2008}}=sum_{k=1}^{2007}{25^{1004}over 25^k+25^{1004}} = sum_{k=1}^{2007}{1over 25^{k-1004}+1} =sum_{j=-1003}^{1003}{1over 2^{-j}+1}$$
Now try your pairing strategy.
answered Nov 21 at 18:11
saulspatz
12.9k21327
Write $$sum_{k=1}^{2007} frac{5^{2008}}{25^k+5^{2008}}=sum_{k=1}^{2007}{25^{1004}over 25^k+25^{1004}} = sum_{k=1}^{2007}{1over 25^{k-1004}+1} =sum_{j=-1003}^{1003}{1over 2^{-j}+1}$$
Now try your pairing strategy.
answered Nov 21 at 18:11
saulspatz
12.9k21327
answered Nov 21 at 18:11
saulspatz
12.9k21327
answered Nov 21 at 18:11
saulspatz
12.9k21327
answered Nov 21 at 18:11
saulspatz
12.9k21327
12.9k21327
add a comment |
add a comment |
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What is the "Gaussian method"?
– user587192
Nov 21 at 17:52
gaus method for calculate for sum but i do not know is this good translation
– Marko Škorić
Nov 21 at 17:57
3
math.stackexchange.com/questions/2541131/…
– lab bhattacharjee
Nov 21 at 18:06