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Proving by induction of $n$ that $sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1 $

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 $sumlimits_k=1^n frac 2^kk leq 2^n$

4

1

$begingroup$

$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$

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

share|cite|improve this question

edited 8 mins ago

Asaf Karagila♦

307k33439770

asked 5 hours ago

BrownieBrownie

1927

$endgroup$

add a comment | 

4

1

$begingroup$

$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$

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

share|cite|improve this question

edited 8 mins ago

Asaf Karagila♦

307k33439770

asked 5 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

share|cite|improve this question

edited 8 mins ago

Asaf Karagila♦

307k33439770

asked 5 hours ago

BrownieBrownie

1927

$endgroup$

share|cite|improve this question

edited 8 mins ago

Asaf Karagila♦

307k33439770

asked 5 hours ago

BrownieBrownie

1927

share|cite|improve this question

edited 8 mins ago

Asaf Karagila♦

307k33439770

asked 5 hours ago

BrownieBrownie

1927

edited 8 mins ago

Asaf Karagila♦

307k33439770

edited 8 mins ago

Asaf Karagila♦

307k33439770

asked 5 hours ago

BrownieBrownie

1927

asked 5 hours ago

BrownieBrownie

1927

2 Answers
2

active

oldest

votes

3

$begingroup$

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.

share|cite|improve this answer

answered 5 hours ago

John Wayland BalesJohn Wayland Bales

15.1k21238

$endgroup$

add a comment | 

2

$begingroup$

Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$

share|cite|improve this answer

answered 4 hours ago

robjohn♦robjohn

270k27312639

$endgroup$

add a comment | 

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3

$begingroup$

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.

share|cite|improve this answer

answered 5 hours ago

John Wayland BalesJohn Wayland Bales

15.1k21238

$endgroup$

add a comment | 

3

$begingroup$

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.

share|cite|improve this answer

answered 5 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.

share|cite|improve this answer

answered 5 hours ago

John Wayland BalesJohn Wayland Bales

15.1k21238

$endgroup$

share|cite|improve this answer

answered 5 hours ago

John Wayland BalesJohn Wayland Bales

15.1k21238

answered 5 hours ago

John Wayland BalesJohn Wayland Bales

15.1k21238

answered 5 hours ago

John Wayland BalesJohn Wayland Bales

15.1k21238

2

$begingroup$

Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$

share|cite|improve this answer

answered 4 hours ago

robjohn♦robjohn

270k27312639

$endgroup$

add a comment | 

2

$begingroup$

Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$

share|cite|improve this answer

answered 4 hours ago

robjohn♦robjohn

270k27312639

$endgroup$

add a comment | 

$begingroup$

Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$

share|cite|improve this answer

answered 4 hours ago

robjohn♦robjohn

270k27312639

$endgroup$

share|cite|improve this answer

answered 4 hours ago

robjohn♦robjohn

270k27312639

answered 4 hours ago

robjohn♦robjohn

270k27312639

answered 4 hours ago

robjohn♦robjohn

270k27312639