Mixing
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Prove that $sigma_n - e$ is decreasing faster than $e-S_n$ given specific $sigma_n$ and $S_n$.
$begingroup$
Let $sigma_n$ and $S_n$ be defined as:
$$
sigma_n = 3 - sum_{k=1}^nfrac{1}{k(k+1)(k+1)!} \
S_n = 1 + sum_{k=1}^nfrac{1}{k!}
$$
Show that $sigma_n - e$ is decreasing faster than $e-S_n$.
I may use anything before the definition of a derivative.
Based on the question we want to eventually show that:
$$
sigma_n - e le e - S_n
$$
Lets adjust $sigma_n$. Define $a_n$:
$$
a_n = sum_{k=1}^nfrac{1}{k(k+1)(k+1)!}
$$
This may be expanded by partial fractions:
$$
begin{align}
a_n &= sum_{k=1}^nfrac{1}{k(k+1)(k+1)!} \
&= sum_{k=1}^nfrac{1}{k(k+1)}cdotfrac{1}{(k+1)!} \
&= sum_{k=1}^nleft(frac{1}{k} - frac{1}{k+1}right)cdotfrac{1}{(k+1)!} \
&= sum_{k=1}^nleft(frac{1}{k} - frac{1}{k+1} - frac{1}{(k+1)^2}right)cdotfrac{1}{k!} \
&= sum_{k=1}^n frac{1}{kk!} - color{red}{sum_{k=1}^n frac{1}{(k+1)k!}} - color{blue}{sum_{k=1}^n frac{1}{(k+1)^2k!}} \
&= sum_{k=1}^n frac{1}{kk!} - color{blue}{sum_{k=1}^n frac{1}{(k+1)(k+1)!}} - color{red}{sum_{k=1}^n frac{1}{(k+1)k!}} \
text{(telescoping)} &= 1 - frac{1}{(n+1)(n+1)!} - sum_{k=1}^n frac{1}{(k+1)!}
end{align}
$$
Now going back to the inequality:
$$
sigma_n - e le e - S_n iff \
sigma_n + S_n le 2e iff \
3 - a_n + S_n le 2e
$$
Replacing the terms with actual sums one may obtain:
$$
begin{align*}
3 - a_n + S_n &= 3 - left(1 - frac{1}{(n+1)(n+1)!} - sum_{k=1}^n frac{1}{(k+1)!}right) + 1 + sum_{k=1}^n{1over k!} tag{1.1} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=1}^n frac{1}{(k+1)!} + sum_{k=1}^n{1over k!} tag{1.2} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=2}^{n+1} frac{1}{k!} + sum_{k=1}^n{1over k!} tag{1.3} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=1}^{n} frac{1}{k!} - 1 + {1over (n+1)!} + sum_{k=1}^n{1over k!} tag{1.4} \
&= 2 + frac{1}{(n+1)(n+1)!} + frac{1}{(n+1)!} + 2sum_{k=1}^{n} frac{1}{k!} tag{1.5} \
&= 2left(1 + frac{1}{2(n+1)!}left({1over n+1} + 1right) + sum_{k=1}^{n} frac{1}{k!}right) tag{1.6} \
&= 2left(frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!}right) tag{1.7}
end{align*}
$$
Which eventually results into:
$$
frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!} le e
$$
Now based on this question:
$$
e - S_n le frac{n+2}{(n+1)(n+1)!}
$$
While I wanted to show:
$$
frac{n+2}{color{red}{2}(n+1)(n+1)!} + S_n le e iff \
frac{n+2}{color{red}{2}(n+1)(n+1)!} le e - S_n
$$
Or summarizing:
$$
frac{n+2}{2(n+1)(n+1)!} le e - S_n le frac{n+2}{(n+1)(n+1)!}
$$
Which seems to be the case. Now I got stuck. How do I proceed from here?
Description of the steps:
$(1.1)$ - replace $sigma_n$ and $S_n$ with sums
$(1.2)$ - cancel $-1+1$
$(1.3)$ - change indexing in the sum
$(1.4)$ - add and subtract $1$ and change the index of the sum. Fetch last term of the sum.
$(1.5)$ - add the sums.
$(1.6)$ - factor out $2$. Factor out $1over (n+1)!$
$(1.7)$ - inject $1$ into the sum and change indexing. Cast brackets to a single fraction
calculus sequences-and-series proof-verification summation
asked Jan 24 at 15:39
romanroman
2,36421224
$endgroup$
add a comment |
$begingroup$
Let $sigma_n$ and $S_n$ be defined as:
$$
sigma_n = 3 - sum_{k=1}^nfrac{1}{k(k+1)(k+1)!} \
S_n = 1 + sum_{k=1}^nfrac{1}{k!}
$$
Show that $sigma_n - e$ is decreasing faster than $e-S_n$.
I may use anything before the definition of a derivative.
Based on the question we want to eventually show that:
$$
sigma_n - e le e - S_n
$$
Lets adjust $sigma_n$. Define $a_n$:
$$
a_n = sum_{k=1}^nfrac{1}{k(k+1)(k+1)!}
$$
This may be expanded by partial fractions:
$$
begin{align}
a_n &= sum_{k=1}^nfrac{1}{k(k+1)(k+1)!} \
&= sum_{k=1}^nfrac{1}{k(k+1)}cdotfrac{1}{(k+1)!} \
&= sum_{k=1}^nleft(frac{1}{k} - frac{1}{k+1}right)cdotfrac{1}{(k+1)!} \
&= sum_{k=1}^nleft(frac{1}{k} - frac{1}{k+1} - frac{1}{(k+1)^2}right)cdotfrac{1}{k!} \
&= sum_{k=1}^n frac{1}{kk!} - color{red}{sum_{k=1}^n frac{1}{(k+1)k!}} - color{blue}{sum_{k=1}^n frac{1}{(k+1)^2k!}} \
&= sum_{k=1}^n frac{1}{kk!} - color{blue}{sum_{k=1}^n frac{1}{(k+1)(k+1)!}} - color{red}{sum_{k=1}^n frac{1}{(k+1)k!}} \
text{(telescoping)} &= 1 - frac{1}{(n+1)(n+1)!} - sum_{k=1}^n frac{1}{(k+1)!}
end{align}
$$
Now going back to the inequality:
$$
sigma_n - e le e - S_n iff \
sigma_n + S_n le 2e iff \
3 - a_n + S_n le 2e
$$
Replacing the terms with actual sums one may obtain:
$$
begin{align*}
3 - a_n + S_n &= 3 - left(1 - frac{1}{(n+1)(n+1)!} - sum_{k=1}^n frac{1}{(k+1)!}right) + 1 + sum_{k=1}^n{1over k!} tag{1.1} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=1}^n frac{1}{(k+1)!} + sum_{k=1}^n{1over k!} tag{1.2} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=2}^{n+1} frac{1}{k!} + sum_{k=1}^n{1over k!} tag{1.3} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=1}^{n} frac{1}{k!} - 1 + {1over (n+1)!} + sum_{k=1}^n{1over k!} tag{1.4} \
&= 2 + frac{1}{(n+1)(n+1)!} + frac{1}{(n+1)!} + 2sum_{k=1}^{n} frac{1}{k!} tag{1.5} \
&= 2left(1 + frac{1}{2(n+1)!}left({1over n+1} + 1right) + sum_{k=1}^{n} frac{1}{k!}right) tag{1.6} \
&= 2left(frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!}right) tag{1.7}
end{align*}
$$
Which eventually results into:
$$
frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!} le e
$$
Now based on this question:
$$
e - S_n le frac{n+2}{(n+1)(n+1)!}
$$
While I wanted to show:
$$
frac{n+2}{color{red}{2}(n+1)(n+1)!} + S_n le e iff \
frac{n+2}{color{red}{2}(n+1)(n+1)!} le e - S_n
$$
Or summarizing:
$$
frac{n+2}{2(n+1)(n+1)!} le e - S_n le frac{n+2}{(n+1)(n+1)!}
$$
Which seems to be the case. Now I got stuck. How do I proceed from here?
Description of the steps:
$(1.1)$ - replace $sigma_n$ and $S_n$ with sums
$(1.2)$ - cancel $-1+1$
$(1.3)$ - change indexing in the sum
$(1.4)$ - add and subtract $1$ and change the index of the sum. Fetch last term of the sum.
$(1.5)$ - add the sums.
$(1.6)$ - factor out $2$. Factor out $1over (n+1)!$
$(1.7)$ - inject $1$ into the sum and change indexing. Cast brackets to a single fraction
calculus sequences-and-series proof-verification summation
asked Jan 24 at 15:39
romanroman
2,36421224
$endgroup$
add a comment |
$begingroup$
Let $sigma_n$ and $S_n$ be defined as:
$$
sigma_n = 3 - sum_{k=1}^nfrac{1}{k(k+1)(k+1)!} \
S_n = 1 + sum_{k=1}^nfrac{1}{k!}
$$
Show that $sigma_n - e$ is decreasing faster than $e-S_n$.
I may use anything before the definition of a derivative.
Based on the question we want to eventually show that:
$$
sigma_n - e le e - S_n
$$
Lets adjust $sigma_n$. Define $a_n$:
$$
a_n = sum_{k=1}^nfrac{1}{k(k+1)(k+1)!}
$$
This may be expanded by partial fractions:
$$
begin{align}
a_n &= sum_{k=1}^nfrac{1}{k(k+1)(k+1)!} \
&= sum_{k=1}^nfrac{1}{k(k+1)}cdotfrac{1}{(k+1)!} \
&= sum_{k=1}^nleft(frac{1}{k} - frac{1}{k+1}right)cdotfrac{1}{(k+1)!} \
&= sum_{k=1}^nleft(frac{1}{k} - frac{1}{k+1} - frac{1}{(k+1)^2}right)cdotfrac{1}{k!} \
&= sum_{k=1}^n frac{1}{kk!} - color{red}{sum_{k=1}^n frac{1}{(k+1)k!}} - color{blue}{sum_{k=1}^n frac{1}{(k+1)^2k!}} \
&= sum_{k=1}^n frac{1}{kk!} - color{blue}{sum_{k=1}^n frac{1}{(k+1)(k+1)!}} - color{red}{sum_{k=1}^n frac{1}{(k+1)k!}} \
text{(telescoping)} &= 1 - frac{1}{(n+1)(n+1)!} - sum_{k=1}^n frac{1}{(k+1)!}
end{align}
$$
Now going back to the inequality:
$$
sigma_n - e le e - S_n iff \
sigma_n + S_n le 2e iff \
3 - a_n + S_n le 2e
$$
Replacing the terms with actual sums one may obtain:
$$
begin{align*}
3 - a_n + S_n &= 3 - left(1 - frac{1}{(n+1)(n+1)!} - sum_{k=1}^n frac{1}{(k+1)!}right) + 1 + sum_{k=1}^n{1over k!} tag{1.1} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=1}^n frac{1}{(k+1)!} + sum_{k=1}^n{1over k!} tag{1.2} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=2}^{n+1} frac{1}{k!} + sum_{k=1}^n{1over k!} tag{1.3} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=1}^{n} frac{1}{k!} - 1 + {1over (n+1)!} + sum_{k=1}^n{1over k!} tag{1.4} \
&= 2 + frac{1}{(n+1)(n+1)!} + frac{1}{(n+1)!} + 2sum_{k=1}^{n} frac{1}{k!} tag{1.5} \
&= 2left(1 + frac{1}{2(n+1)!}left({1over n+1} + 1right) + sum_{k=1}^{n} frac{1}{k!}right) tag{1.6} \
&= 2left(frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!}right) tag{1.7}
end{align*}
$$
Which eventually results into:
$$
frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!} le e
$$
Now based on this question:
$$
e - S_n le frac{n+2}{(n+1)(n+1)!}
$$
While I wanted to show:
$$
frac{n+2}{color{red}{2}(n+1)(n+1)!} + S_n le e iff \
frac{n+2}{color{red}{2}(n+1)(n+1)!} le e - S_n
$$
Or summarizing:
$$
frac{n+2}{2(n+1)(n+1)!} le e - S_n le frac{n+2}{(n+1)(n+1)!}
$$
Which seems to be the case. Now I got stuck. How do I proceed from here?
Description of the steps:
$(1.1)$ - replace $sigma_n$ and $S_n$ with sums
$(1.2)$ - cancel $-1+1$
$(1.3)$ - change indexing in the sum
$(1.4)$ - add and subtract $1$ and change the index of the sum. Fetch last term of the sum.
$(1.5)$ - add the sums.
$(1.6)$ - factor out $2$. Factor out $1over (n+1)!$
$(1.7)$ - inject $1$ into the sum and change indexing. Cast brackets to a single fraction
calculus sequences-and-series proof-verification summation
asked Jan 24 at 15:39
romanroman
2,36421224
$endgroup$
Let $sigma_n$ and $S_n$ be defined as:
$$
sigma_n = 3 - sum_{k=1}^nfrac{1}{k(k+1)(k+1)!} \
S_n = 1 + sum_{k=1}^nfrac{1}{k!}
$$
Show that $sigma_n - e$ is decreasing faster than $e-S_n$.
I may use anything before the definition of a derivative.
Based on the question we want to eventually show that:
$$
sigma_n - e le e - S_n
$$
Lets adjust $sigma_n$. Define $a_n$:
$$
a_n = sum_{k=1}^nfrac{1}{k(k+1)(k+1)!}
$$
This may be expanded by partial fractions:
$$
begin{align}
a_n &= sum_{k=1}^nfrac{1}{k(k+1)(k+1)!} \
&= sum_{k=1}^nfrac{1}{k(k+1)}cdotfrac{1}{(k+1)!} \
&= sum_{k=1}^nleft(frac{1}{k} - frac{1}{k+1}right)cdotfrac{1}{(k+1)!} \
&= sum_{k=1}^nleft(frac{1}{k} - frac{1}{k+1} - frac{1}{(k+1)^2}right)cdotfrac{1}{k!} \
&= sum_{k=1}^n frac{1}{kk!} - color{red}{sum_{k=1}^n frac{1}{(k+1)k!}} - color{blue}{sum_{k=1}^n frac{1}{(k+1)^2k!}} \
&= sum_{k=1}^n frac{1}{kk!} - color{blue}{sum_{k=1}^n frac{1}{(k+1)(k+1)!}} - color{red}{sum_{k=1}^n frac{1}{(k+1)k!}} \
text{(telescoping)} &= 1 - frac{1}{(n+1)(n+1)!} - sum_{k=1}^n frac{1}{(k+1)!}
end{align}
$$
Now going back to the inequality:
$$
sigma_n - e le e - S_n iff \
sigma_n + S_n le 2e iff \
3 - a_n + S_n le 2e
$$
Replacing the terms with actual sums one may obtain:
$$
begin{align*}
3 - a_n + S_n &= 3 - left(1 - frac{1}{(n+1)(n+1)!} - sum_{k=1}^n frac{1}{(k+1)!}right) + 1 + sum_{k=1}^n{1over k!} tag{1.1} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=1}^n frac{1}{(k+1)!} + sum_{k=1}^n{1over k!} tag{1.2} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=2}^{n+1} frac{1}{k!} + sum_{k=1}^n{1over k!} tag{1.3} \
&= 3 + frac{1}{(n+1)(n+1)!} + sum_{k=1}^{n} frac{1}{k!} - 1 + {1over (n+1)!} + sum_{k=1}^n{1over k!} tag{1.4} \
&= 2 + frac{1}{(n+1)(n+1)!} + frac{1}{(n+1)!} + 2sum_{k=1}^{n} frac{1}{k!} tag{1.5} \
&= 2left(1 + frac{1}{2(n+1)!}left({1over n+1} + 1right) + sum_{k=1}^{n} frac{1}{k!}right) tag{1.6} \
&= 2left(frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!}right) tag{1.7}
end{align*}
$$
Which eventually results into:
$$
frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!} le e
$$
Now based on this question:
$$
e - S_n le frac{n+2}{(n+1)(n+1)!}
$$
While I wanted to show:
$$
frac{n+2}{color{red}{2}(n+1)(n+1)!} + S_n le e iff \
frac{n+2}{color{red}{2}(n+1)(n+1)!} le e - S_n
$$
Or summarizing:
$$
frac{n+2}{2(n+1)(n+1)!} le e - S_n le frac{n+2}{(n+1)(n+1)!}
$$
Which seems to be the case. Now I got stuck. How do I proceed from here?
Description of the steps:
$(1.1)$ - replace $sigma_n$ and $S_n$ with sums
$(1.2)$ - cancel $-1+1$
$(1.3)$ - change indexing in the sum
$(1.4)$ - add and subtract $1$ and change the index of the sum. Fetch last term of the sum.
$(1.5)$ - add the sums.
$(1.6)$ - factor out $2$. Factor out $1over (n+1)!$
$(1.7)$ - inject $1$ into the sum and change indexing. Cast brackets to a single fraction
calculus sequences-and-series proof-verification summation
calculus sequences-and-series proof-verification summation
asked Jan 24 at 15:39
romanroman
2,36421224
asked Jan 24 at 15:39
romanroman
2,36421224
edited Jan 24 at 22:08
roman
asked Jan 24 at 15:39
romanroman
2,36421224
asked Jan 24 at 15:39
romanroman
2,36421224
asked Jan 24 at 15:39
romanroman
2,36421224
2,36421224
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Presuming my cursory review of your work leading up to the inequality $frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!} le e$ didn't miss an error and assuming $n in mathbb N$, we may proceed as follows:
begin{aligned}\
e-sum_{k=0}^{n} frac{1}{k!}
&= sum_{k=n+1}^{infty} {1over k!}\
&geq frac1{(n+1)!}+frac1{(n+2)!}\
&geq frac1{2(n+1)!}+frac1{(2n+2)(n+1)!} && (n+2leq2n+2) \
&=frac{n+1}{2(n+1)(n+1)!}+frac1{2(n+1)(n+1)!} \
&=frac{n+2}{2(n+1)(n+1)!}
end{aligned}
Seems to me that you did most of the work in your original question.
answered Jan 24 at 23:02
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
1
$begingroup$
Wonderful. Thank you!
$endgroup$
– roman
Jan 24 at 23:10
add a comment |
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$begingroup$
Presuming my cursory review of your work leading up to the inequality $frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!} le e$ didn't miss an error and assuming $n in mathbb N$, we may proceed as follows:
begin{aligned}\
e-sum_{k=0}^{n} frac{1}{k!}
&= sum_{k=n+1}^{infty} {1over k!}\
&geq frac1{(n+1)!}+frac1{(n+2)!}\
&geq frac1{2(n+1)!}+frac1{(2n+2)(n+1)!} && (n+2leq2n+2) \
&=frac{n+1}{2(n+1)(n+1)!}+frac1{2(n+1)(n+1)!} \
&=frac{n+2}{2(n+1)(n+1)!}
end{aligned}
Seems to me that you did most of the work in your original question.
answered Jan 24 at 23:02
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
1
$begingroup$
Wonderful. Thank you!
$endgroup$
– roman
Jan 24 at 23:10
add a comment |
$begingroup$
Presuming my cursory review of your work leading up to the inequality $frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!} le e$ didn't miss an error and assuming $n in mathbb N$, we may proceed as follows:
begin{aligned}\
e-sum_{k=0}^{n} frac{1}{k!}
&= sum_{k=n+1}^{infty} {1over k!}\
&geq frac1{(n+1)!}+frac1{(n+2)!}\
&geq frac1{2(n+1)!}+frac1{(2n+2)(n+1)!} && (n+2leq2n+2) \
&=frac{n+1}{2(n+1)(n+1)!}+frac1{2(n+1)(n+1)!} \
&=frac{n+2}{2(n+1)(n+1)!}
end{aligned}
Seems to me that you did most of the work in your original question.
answered Jan 24 at 23:02
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
1
$begingroup$
Wonderful. Thank you!
$endgroup$
– roman
Jan 24 at 23:10
add a comment |
$begingroup$
Presuming my cursory review of your work leading up to the inequality $frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!} le e$ didn't miss an error and assuming $n in mathbb N$, we may proceed as follows:
begin{aligned}\
e-sum_{k=0}^{n} frac{1}{k!}
&= sum_{k=n+1}^{infty} {1over k!}\
&geq frac1{(n+1)!}+frac1{(n+2)!}\
&geq frac1{2(n+1)!}+frac1{(2n+2)(n+1)!} && (n+2leq2n+2) \
&=frac{n+1}{2(n+1)(n+1)!}+frac1{2(n+1)(n+1)!} \
&=frac{n+2}{2(n+1)(n+1)!}
end{aligned}
Seems to me that you did most of the work in your original question.
answered Jan 24 at 23:02
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
Presuming my cursory review of your work leading up to the inequality $frac{(n+2)}{2(n+1)(n+1)!} + sum_{k=0}^{n} frac{1}{k!} le e$ didn't miss an error and assuming $n in mathbb N$, we may proceed as follows:
begin{aligned}\
e-sum_{k=0}^{n} frac{1}{k!}
&= sum_{k=n+1}^{infty} {1over k!}\
&geq frac1{(n+1)!}+frac1{(n+2)!}\
&geq frac1{2(n+1)!}+frac1{(2n+2)(n+1)!} && (n+2leq2n+2) \
&=frac{n+1}{2(n+1)(n+1)!}+frac1{2(n+1)(n+1)!} \
&=frac{n+2}{2(n+1)(n+1)!}
end{aligned}
Seems to me that you did most of the work in your original question.
answered Jan 24 at 23:02
Matt A PeltoMatt A Pelto
2,602621
edited Jan 24 at 23:07
answered Jan 24 at 23:02
Matt A PeltoMatt A Pelto
2,602621
answered Jan 24 at 23:02
Matt A PeltoMatt A Pelto
2,602621
answered Jan 24 at 23:02
Matt A PeltoMatt A Pelto
2,602621
2,602621
1
$begingroup$
Wonderful. Thank you!
$endgroup$
– roman
Jan 24 at 23:10
add a comment |
1
$begingroup$
Wonderful. Thank you!
$endgroup$
– roman
Jan 24 at 23:10
1
1
$begingroup$
Wonderful. Thank you!
$endgroup$
– roman
Jan 24 at 23:10
$begingroup$
Wonderful. Thank you!
$endgroup$
– roman
Jan 24 at 23:10
add a comment |
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