Proving by induction of n. Is this correct until this point?
$begingroup$
$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$
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
asked 4 hours ago
BrownieBrownie
1927
$endgroup$
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
asked 4 hours ago
BrownieBrownie
1927
$endgroup$
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
asked 4 hours ago
BrownieBrownie
1927
$endgroup$
$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$
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
asked 4 hours ago
BrownieBrownie
1927
asked 4 hours ago
BrownieBrownie
1927
asked 4 hours ago
BrownieBrownie
1927
asked 4 hours ago
BrownieBrownie
1927
asked 4 hours ago
BrownieBrownie
1927
1927
add a comment |
add a comment |
2 Answers
2
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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 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
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 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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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 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
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 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
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 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
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 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
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 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
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 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
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 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
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 3 hours ago
robjohn♦robjohn
270k27312639
answered 3 hours ago
robjohn♦robjohn
270k27312639
answered 3 hours ago
robjohn♦robjohn
270k27312639
answered 3 hours ago
robjohn♦robjohn
270k27312639
270k27312639
add a comment |
add a comment |
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