Proving by induction of n. Is this correct until this point?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 $sumlimits_k=1^n frac kk+1 leq n - frac1n+1$Prove by induction on n that $sumlimits_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1$Prove by induction on n that $sumlimits_k=1^n frac 2^kk leq 2^n$
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Proving by induction of n. Is this correct until this point?
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 $sumlimits_k=1^n frac kk+1 leq n - frac1n+1$Prove by induction on n that $sumlimits_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1$Prove by induction on n that $sumlimits_k=1^n frac 2^kk leq 2^n$
$begingroup$
$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_k=1^n frac k+2k(k+1)2^k+1 = frac38$$
RHS-
$$frac12 - frac1(n+1)2^n+1 = frac38$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_k=1^n+1 frac k+2k(k+1)2^k+1 +frac n+3(n+1)(n+2)2^n+2$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac12 - frac1(n+1)2^n+1 +frac n+3(n+1)(n+2)2^n+2
$$
and then I broke up $frac n+3(n+1)(n+2)2^n+2$ into
$$frac2(n+2)-(n+1)(n+1)(n+2)2^n+2$$
$$=$$
$$frac2(n+1)2^n+2 - frac1(n+2)2^n+2$$
$$=$$
$$frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then put it back in with the rest of the equation, bringing me to
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then
$$frac12 -frac2(n+1)2^n+1 - frac1(n+2)2^n+2$$
and
$$frac12 -frac1(n+1)2^n - frac1(n+2)2^n+2$$
$$frac12 -frac(n+2)2^n+2 - (n+1)2^n(n+1)(n+2)2^2n+2 $$
which I think simplifies down to this after factoring out a $2^n$ from the numerator?
$$frac12 -frac2^n((n+2)2^2 - (n+1))(n+1)(n+2)2^2n+2 $$
canceling out $2^n$
$$frac12 -frac(3n-7)(n+1)(n+2)2^n+2 $$
and I'm stuck, please help!
discrete-mathematics induction
asked 4 hours ago
BrownieBrownie
1927
$endgroup$
add a comment |
$begingroup$
$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_k=1^n frac k+2k(k+1)2^k+1 = frac38$$
RHS-
$$frac12 - frac1(n+1)2^n+1 = frac38$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_k=1^n+1 frac k+2k(k+1)2^k+1 +frac n+3(n+1)(n+2)2^n+2$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac12 - frac1(n+1)2^n+1 +frac n+3(n+1)(n+2)2^n+2
$$
and then I broke up $frac n+3(n+1)(n+2)2^n+2$ into
$$frac2(n+2)-(n+1)(n+1)(n+2)2^n+2$$
$$=$$
$$frac2(n+1)2^n+2 - frac1(n+2)2^n+2$$
$$=$$
$$frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then put it back in with the rest of the equation, bringing me to
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then
$$frac12 -frac2(n+1)2^n+1 - frac1(n+2)2^n+2$$
and
$$frac12 -frac1(n+1)2^n - frac1(n+2)2^n+2$$
$$frac12 -frac(n+2)2^n+2 - (n+1)2^n(n+1)(n+2)2^2n+2 $$
which I think simplifies down to this after factoring out a $2^n$ from the numerator?
$$frac12 -frac2^n((n+2)2^2 - (n+1))(n+1)(n+2)2^2n+2 $$
canceling out $2^n$
$$frac12 -frac(3n-7)(n+1)(n+2)2^n+2 $$
and I'm stuck, please help!
discrete-mathematics induction
asked 4 hours ago
BrownieBrownie
1927
$endgroup$
add a comment |
$begingroup$
$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_k=1^n frac k+2k(k+1)2^k+1 = frac38$$
RHS-
$$frac12 - frac1(n+1)2^n+1 = frac38$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_k=1^n+1 frac k+2k(k+1)2^k+1 +frac n+3(n+1)(n+2)2^n+2$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac12 - frac1(n+1)2^n+1 +frac n+3(n+1)(n+2)2^n+2
$$
and then I broke up $frac n+3(n+1)(n+2)2^n+2$ into
$$frac2(n+2)-(n+1)(n+1)(n+2)2^n+2$$
$$=$$
$$frac2(n+1)2^n+2 - frac1(n+2)2^n+2$$
$$=$$
$$frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then put it back in with the rest of the equation, bringing me to
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then
$$frac12 -frac2(n+1)2^n+1 - frac1(n+2)2^n+2$$
and
$$frac12 -frac1(n+1)2^n - frac1(n+2)2^n+2$$
$$frac12 -frac(n+2)2^n+2 - (n+1)2^n(n+1)(n+2)2^2n+2 $$
which I think simplifies down to this after factoring out a $2^n$ from the numerator?
$$frac12 -frac2^n((n+2)2^2 - (n+1))(n+1)(n+2)2^2n+2 $$
canceling out $2^n$
$$frac12 -frac(3n-7)(n+1)(n+2)2^n+2 $$
and I'm stuck, please help!
discrete-mathematics induction
asked 4 hours ago
BrownieBrownie
1927
$endgroup$
$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_k=1^n frac k+2k(k+1)2^k+1 = frac38$$
RHS-
$$frac12 - frac1(n+1)2^n+1 = frac38$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_k=1^n+1 frac k+2k(k+1)2^k+1 +frac n+3(n+1)(n+2)2^n+2$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac12 - frac1(n+1)2^n+1 +frac n+3(n+1)(n+2)2^n+2
$$
and then I broke up $frac n+3(n+1)(n+2)2^n+2$ into
$$frac2(n+2)-(n+1)(n+1)(n+2)2^n+2$$
$$=$$
$$frac2(n+1)2^n+2 - frac1(n+2)2^n+2$$
$$=$$
$$frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then put it back in with the rest of the equation, bringing me to
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then
$$frac12 -frac2(n+1)2^n+1 - frac1(n+2)2^n+2$$
and
$$frac12 -frac1(n+1)2^n - frac1(n+2)2^n+2$$
$$frac12 -frac(n+2)2^n+2 - (n+1)2^n(n+1)(n+2)2^2n+2 $$
which I think simplifies down to this after factoring out a $2^n$ from the numerator?
$$frac12 -frac2^n((n+2)2^2 - (n+1))(n+1)(n+2)2^2n+2 $$
canceling out $2^n$
$$frac12 -frac(3n-7)(n+1)(n+2)2^n+2 $$
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
active
oldest
votes
$begingroup$
Your error is just after the sixth step from the bottom:
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2=frac12 -frac1(n+2)2^n+2$$
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
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$
answered 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your error is just after the sixth step from the bottom:
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2=frac12 -frac1(n+2)2^n+2$$
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:
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2=frac12 -frac1(n+2)2^n+2$$
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:
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2=frac12 -frac1(n+2)2^n+2$$
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:
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2=frac12 -frac1(n+2)2^n+2$$
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
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$
answered 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$
answered 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$
answered 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$
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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