Proving by induction of $n$ that $sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}} $...
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Proving by induction of $n$ that $sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}} $
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)prove inequality by induction — Discrete mathProve $25^n>6^n$ using inductionTrying to simplify an expression for an induction proof.Induction on summation inequality stuck on Induction stepProve by Induction: Summation of Factorial (n! * n)Prove that $n! > n^{3}$ for every integer $n ge 6$ using inductionProving by induction on $n$ that $sum limits_{k=1}^n (k+1)2^{k} = n2^{n+1} $5. Prove by induction on $n$ that $sumlimits_{k=1}^n frac k{k+1} leq n - frac1{n+1}$Prove by induction on n that $sumlimits_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}$Prove by induction on n that $sumlimits_{k=1}^n frac {2^{k}}{k} leq 2^{n}$
$begingroup$
$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$
RHS-
$$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then
$$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
and
$$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$
$$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
discrete-mathematics induction
edited Mar 26 at 6:28
Asaf Karagila♦
309k33441775
asked Mar 26 at 1:04
BrownieBrownie
3467
$endgroup$
add a comment |
$begingroup$
$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$
RHS-
$$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then
$$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
and
$$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$
$$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
discrete-mathematics induction
edited Mar 26 at 6:28
Asaf Karagila♦
309k33441775
asked Mar 26 at 1:04
BrownieBrownie
3467
$endgroup$
add a comment |
$begingroup$
$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$
RHS-
$$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then
$$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
and
$$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$
$$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
discrete-mathematics induction
edited Mar 26 at 6:28
Asaf Karagila♦
309k33441775
asked Mar 26 at 1:04
BrownieBrownie
3467
$endgroup$
$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$
RHS-
$$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then
$$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
and
$$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$
$$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
discrete-mathematics induction
discrete-mathematics induction
edited Mar 26 at 6:28
Asaf Karagila♦
309k33441775
asked Mar 26 at 1:04
BrownieBrownie
3467
edited Mar 26 at 6:28
Asaf Karagila♦
309k33441775
asked Mar 26 at 1:04
BrownieBrownie
3467
edited Mar 26 at 6:28
Asaf Karagila♦
309k33441775
edited Mar 26 at 6:28
Asaf Karagila♦
309k33441775
edited Mar 26 at 6:28
Asaf Karagila♦
309k33441775
309k33441775
asked Mar 26 at 1:04
BrownieBrownie
3467
asked Mar 26 at 1:04
BrownieBrownie
3467
asked Mar 26 at 1:04
BrownieBrownie
3467
3467
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your error is just after the sixth step from the bottom:
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered Mar 26 at 1:24
John Wayland BalesJohn Wayland Bales
15.3k21238
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
begin{align}
sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
&=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
&=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
&=frac12-frac1{(n+1)2^{n+1}}
end{align}
$$
answered Mar 26 at 2:17
robjohn♦robjohn
271k27316643
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
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$begingroup$
Your error is just after the sixth step from the bottom:
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered Mar 26 at 1:24
John Wayland BalesJohn Wayland Bales
15.3k21238
$endgroup$
add a comment |
$begingroup$
Your error is just after the sixth step from the bottom:
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered Mar 26 at 1:24
John Wayland BalesJohn Wayland Bales
15.3k21238
$endgroup$
add a comment |
$begingroup$
Your error is just after the sixth step from the bottom:
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered Mar 26 at 1:24
John Wayland BalesJohn Wayland Bales
15.3k21238
$endgroup$
Your error is just after the sixth step from the bottom:
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered Mar 26 at 1:24
John Wayland BalesJohn Wayland Bales
15.3k21238
answered Mar 26 at 1:24
John Wayland BalesJohn Wayland Bales
15.3k21238
answered Mar 26 at 1:24
John Wayland BalesJohn Wayland Bales
15.3k21238
answered Mar 26 at 1:24
John Wayland BalesJohn Wayland Bales
15.3k21238
15.3k21238
add a comment |
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
begin{align}
sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
&=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
&=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
&=frac12-frac1{(n+1)2^{n+1}}
end{align}
$$
answered Mar 26 at 2:17
robjohn♦robjohn
271k27316643
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
begin{align}
sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
&=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
&=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
&=frac12-frac1{(n+1)2^{n+1}}
end{align}
$$
answered Mar 26 at 2:17
robjohn♦robjohn
271k27316643
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
begin{align}
sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
&=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
&=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
&=frac12-frac1{(n+1)2^{n+1}}
end{align}
$$
answered Mar 26 at 2:17
robjohn♦robjohn
271k27316643
$endgroup$
Using a telescoping sum, we get
$$
begin{align}
sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
&=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
&=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
&=frac12-frac1{(n+1)2^{n+1}}
end{align}
$$
answered Mar 26 at 2:17
robjohn♦robjohn
271k27316643
answered Mar 26 at 2:17
robjohn♦robjohn
271k27316643
answered Mar 26 at 2:17
robjohn♦robjohn
271k27316643
answered Mar 26 at 2:17
robjohn♦robjohn
271k27316643
271k27316643
add a comment |
add a comment |
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