Mixing
Finding $sum_i=1^100 a_i$ given that $sqrta_1+sqrta_2-1+sqrta_3-2+dots+sqrta_n-(n-1)=frac12(a_1+a_2+dots+a_n)=fracn(n-3)4$
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Let $a_1,a_2,dots,a_n$ be real numbers such that
$$sqrta_1+sqrta_2-1+sqrta_3-2+dots+sqrta_n-(n-1)=frac12(a_1+a_2+dots+a_n)=fracn(n-3)4$$
Compute the value of $sum_i=1^100 a_i$.
Can't I just use simply equation (i) to get $sum_i=1^100a_i = frac(n)(n-3)2$ and by putting $n=100$ I get its summation = $4850$
Why is it wrong ? please anybody explain me ?
The solution given in the book is as follows:
Let $sqrta_1=b_1$
beginalign*
sqrta_2-1&=b_2 \
sqrta_3-2&=b_3 \
ldots\
sqrta_n-(n-1)&=b_n \
endalign*
beginalign*
therefore
& b_1 + b_2 + dots + b_n = \
& frac12 left[b_1^2+(b_2^2+1)+dots+(b_n^2-(n-1))right] - fracn(n-3)4
endalign*
beginalign*
therefore
& sum b_i = frac12 [(b_1^2+b_2^2+dots+b_n^2)]+ \
& (1+2+3+dots+(n-1))] - fracn(n-3)4 \
Rightarrow
& 2sum b_i = sum b_i^2 + fracn(n-1)2 - fracn(n-3)4 \
Rightarrow
& 2sum b_i = sum b_i^2 + n \
Rightarrow
& sum b_i^2 - 2sum b_i + sum 1 = 0 \
& b_1-1 =0 qquadRightarrowqquad b_1^2=a_1=1 \
& b_2-1=0 qquadRightarrowqquad b_2^2 = a_2-1 = 1 qquadRightarrowqquad a_2=2\
& b_3-1=0 qquadRightarrowqquad b_3^2 = a_3-2 = 1 qquadRightarrowqquad a_3=3
endalign*
and so on. Hence $a_n=n$.
$$therefore sum_i=1^100 a_i = 1+2+3+dots+100=5050.$$
sequences-and-series algebra-precalculus summation
edited Jul 25 at 15:23
Siong Thye Goh
77.4k134795
asked Jul 24 at 16:25
Rafael Nadal
1056
 |Â
show 6 more comments
up vote
-1
down vote
favorite
Let $a_1,a_2,dots,a_n$ be real numbers such that
$$sqrta_1+sqrta_2-1+sqrta_3-2+dots+sqrta_n-(n-1)=frac12(a_1+a_2+dots+a_n)=fracn(n-3)4$$
Compute the value of $sum_i=1^100 a_i$.
Can't I just use simply equation (i) to get $sum_i=1^100a_i = frac(n)(n-3)2$ and by putting $n=100$ I get its summation = $4850$
Why is it wrong ? please anybody explain me ?
The solution given in the book is as follows:
Let $sqrta_1=b_1$
beginalign*
sqrta_2-1&=b_2 \
sqrta_3-2&=b_3 \
ldots\
sqrta_n-(n-1)&=b_n \
endalign*
beginalign*
therefore
& b_1 + b_2 + dots + b_n = \
& frac12 left[b_1^2+(b_2^2+1)+dots+(b_n^2-(n-1))right] - fracn(n-3)4
endalign*
beginalign*
therefore
& sum b_i = frac12 [(b_1^2+b_2^2+dots+b_n^2)]+ \
& (1+2+3+dots+(n-1))] - fracn(n-3)4 \
Rightarrow
& 2sum b_i = sum b_i^2 + fracn(n-1)2 - fracn(n-3)4 \
Rightarrow
& 2sum b_i = sum b_i^2 + n \
Rightarrow
& sum b_i^2 - 2sum b_i + sum 1 = 0 \
& b_1-1 =0 qquadRightarrowqquad b_1^2=a_1=1 \
& b_2-1=0 qquadRightarrowqquad b_2^2 = a_2-1 = 1 qquadRightarrowqquad a_2=2\
& b_3-1=0 qquadRightarrowqquad b_3^2 = a_3-2 = 1 qquadRightarrowqquad a_3=3
endalign*
and so on. Hence $a_n=n$.
$$therefore sum_i=1^100 a_i = 1+2+3+dots+100=5050.$$
sequences-and-series algebra-precalculus summation
edited Jul 25 at 15:23
Siong Thye Goh
77.4k134795
asked Jul 24 at 16:25
Rafael Nadal
1056
1
Looks to me like the textbook is in error. $sum_n=1^100n=5050$ may have something to do with the mistake.
– saulspatz
Jul 24 at 16:35
1
You can type the solution in your question.
– saulspatz
Jul 24 at 16:42
2
No. You should type your question and not post images. If it's not worth your time and trouble to type the question, why is it worth my time and trouble to type an answer?
– saulspatz
Jul 24 at 16:54
2
Read my last comment. Stop sending me messages. I don't want to chat with you.
– saulspatz
Jul 24 at 17:05
2
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have included some text from the pictures into your post. Please, edit it further if needed.
– Martin Sleziak
Jul 25 at 8:42
 |Â
show 6 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $a_1,a_2,dots,a_n$ be real numbers such that
$$sqrta_1+sqrta_2-1+sqrta_3-2+dots+sqrta_n-(n-1)=frac12(a_1+a_2+dots+a_n)=fracn(n-3)4$$
Compute the value of $sum_i=1^100 a_i$.
Can't I just use simply equation (i) to get $sum_i=1^100a_i = frac(n)(n-3)2$ and by putting $n=100$ I get its summation = $4850$
Why is it wrong ? please anybody explain me ?
The solution given in the book is as follows:
Let $sqrta_1=b_1$
beginalign*
sqrta_2-1&=b_2 \
sqrta_3-2&=b_3 \
ldots\
sqrta_n-(n-1)&=b_n \
endalign*
beginalign*
therefore
& b_1 + b_2 + dots + b_n = \
& frac12 left[b_1^2+(b_2^2+1)+dots+(b_n^2-(n-1))right] - fracn(n-3)4
endalign*
beginalign*
therefore
& sum b_i = frac12 [(b_1^2+b_2^2+dots+b_n^2)]+ \
& (1+2+3+dots+(n-1))] - fracn(n-3)4 \
Rightarrow
& 2sum b_i = sum b_i^2 + fracn(n-1)2 - fracn(n-3)4 \
Rightarrow
& 2sum b_i = sum b_i^2 + n \
Rightarrow
& sum b_i^2 - 2sum b_i + sum 1 = 0 \
& b_1-1 =0 qquadRightarrowqquad b_1^2=a_1=1 \
& b_2-1=0 qquadRightarrowqquad b_2^2 = a_2-1 = 1 qquadRightarrowqquad a_2=2\
& b_3-1=0 qquadRightarrowqquad b_3^2 = a_3-2 = 1 qquadRightarrowqquad a_3=3
endalign*
and so on. Hence $a_n=n$.
$$therefore sum_i=1^100 a_i = 1+2+3+dots+100=5050.$$
sequences-and-series algebra-precalculus summation
edited Jul 25 at 15:23
Siong Thye Goh
77.4k134795
asked Jul 24 at 16:25
Rafael Nadal
1056
Let $a_1,a_2,dots,a_n$ be real numbers such that
$$sqrta_1+sqrta_2-1+sqrta_3-2+dots+sqrta_n-(n-1)=frac12(a_1+a_2+dots+a_n)=fracn(n-3)4$$
Compute the value of $sum_i=1^100 a_i$.
Can't I just use simply equation (i) to get $sum_i=1^100a_i = frac(n)(n-3)2$ and by putting $n=100$ I get its summation = $4850$
Why is it wrong ? please anybody explain me ?
The solution given in the book is as follows:
Let $sqrta_1=b_1$
beginalign*
sqrta_2-1&=b_2 \
sqrta_3-2&=b_3 \
ldots\
sqrta_n-(n-1)&=b_n \
endalign*
beginalign*
therefore
& b_1 + b_2 + dots + b_n = \
& frac12 left[b_1^2+(b_2^2+1)+dots+(b_n^2-(n-1))right] - fracn(n-3)4
endalign*
beginalign*
therefore
& sum b_i = frac12 [(b_1^2+b_2^2+dots+b_n^2)]+ \
& (1+2+3+dots+(n-1))] - fracn(n-3)4 \
Rightarrow
& 2sum b_i = sum b_i^2 + fracn(n-1)2 - fracn(n-3)4 \
Rightarrow
& 2sum b_i = sum b_i^2 + n \
Rightarrow
& sum b_i^2 - 2sum b_i + sum 1 = 0 \
& b_1-1 =0 qquadRightarrowqquad b_1^2=a_1=1 \
& b_2-1=0 qquadRightarrowqquad b_2^2 = a_2-1 = 1 qquadRightarrowqquad a_2=2\
& b_3-1=0 qquadRightarrowqquad b_3^2 = a_3-2 = 1 qquadRightarrowqquad a_3=3
endalign*
and so on. Hence $a_n=n$.
$$therefore sum_i=1^100 a_i = 1+2+3+dots+100=5050.$$
sequences-and-series algebra-precalculus summation
edited Jul 25 at 15:23
Siong Thye Goh
77.4k134795
asked Jul 24 at 16:25
Rafael Nadal
1056
edited Jul 25 at 15:23
Siong Thye Goh
77.4k134795
edited Jul 25 at 15:23
Siong Thye Goh
77.4k134795
edited Jul 25 at 15:23
Siong Thye Goh
77.4k134795
77.4k134795
asked Jul 24 at 16:25
Rafael Nadal
1056
asked Jul 24 at 16:25
Rafael Nadal
1056
asked Jul 24 at 16:25
Rafael Nadal
1056
1056
1
Looks to me like the textbook is in error. $sum_n=1^100n=5050$ may have something to do with the mistake.
– saulspatz
Jul 24 at 16:35
1
You can type the solution in your question.
– saulspatz
Jul 24 at 16:42
2
No. You should type your question and not post images. If it's not worth your time and trouble to type the question, why is it worth my time and trouble to type an answer?
– saulspatz
Jul 24 at 16:54
2
Read my last comment. Stop sending me messages. I don't want to chat with you.
– saulspatz
Jul 24 at 17:05
2
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have included some text from the pictures into your post. Please, edit it further if needed.
– Martin Sleziak
Jul 25 at 8:42
 |Â
show 6 more comments
1
Looks to me like the textbook is in error. $sum_n=1^100n=5050$ may have something to do with the mistake.
– saulspatz
Jul 24 at 16:35
1
You can type the solution in your question.
– saulspatz
Jul 24 at 16:42
2
No. You should type your question and not post images. If it's not worth your time and trouble to type the question, why is it worth my time and trouble to type an answer?
– saulspatz
Jul 24 at 16:54
2
Read my last comment. Stop sending me messages. I don't want to chat with you.
– saulspatz
Jul 24 at 17:05
2
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have included some text from the pictures into your post. Please, edit it further if needed.
– Martin Sleziak
Jul 25 at 8:42
1
1
Looks to me like the textbook is in error. $sum_n=1^100n=5050$ may have something to do with the mistake.
– saulspatz
Jul 24 at 16:35
Looks to me like the textbook is in error. $sum_n=1^100n=5050$ may have something to do with the mistake.
– saulspatz
Jul 24 at 16:35
1
1
You can type the solution in your question.
– saulspatz
Jul 24 at 16:42
You can type the solution in your question.
– saulspatz
Jul 24 at 16:42
2
2
No. You should type your question and not post images. If it's not worth your time and trouble to type the question, why is it worth my time and trouble to type an answer?
– saulspatz
Jul 24 at 16:54
No. You should type your question and not post images. If it's not worth your time and trouble to type the question, why is it worth my time and trouble to type an answer?
– saulspatz
Jul 24 at 16:54
2
2
Read my last comment. Stop sending me messages. I don't want to chat with you.
– saulspatz
Jul 24 at 17:05
Read my last comment. Stop sending me messages. I don't want to chat with you.
– saulspatz
Jul 24 at 17:05
2
2
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have included some text from the pictures into your post. Please, edit it further if needed.
– Martin Sleziak
Jul 25 at 8:42
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have included some text from the pictures into your post. Please, edit it further if needed.
– Martin Sleziak
Jul 25 at 8:42
 |Â
show 6 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
You are indeed given that
$$sum_i=1^na_i=fracn(n-3)2$$
and upon substituting $n=100$, we get
$$frac100(97)2=4850$$
It is correct.
Remark:
The correction is the second equality should be a minus sign, from there we can prove that $a_n=n$ and hence the sum of the first $100$ positive integers is $5050$.
That is the actual question is
Let $a_1, a_2, ldots , a_n$ be real numbers such that
beginalign&sqrta_1+sqrta_2-1+sqrta_3-2+ldots + sqrta_n-(n-1) =\
½ (a_1+a_2+ldots+a_n)colorred-fracn(n-3)4endalign
Compute the value of $sum_i=1^100a_i$.
answered Jul 24 at 16:30
Siong Thye Goh
77.4k134795
by this method answer must be 4850 but answer in my textbook is 5050
– Rafael Nadal
Jul 24 at 16:33
which book are you using? regardless of which method, all correct methods should lead to the same solution.
– Siong Thye Goh
Jul 24 at 16:34
sir i can send u snapshot of its solution please clarify my doubt
– Rafael Nadal
Jul 24 at 16:36
1
most likely it is an error in the book solution. there are mistakes in books.
– Siong Thye Goh
Jul 24 at 16:37
sir tell me anyplace where i can send u snapshot of solution given in my book
– Rafael Nadal
Jul 24 at 16:38
 |Â
show 2 more comments
up vote
1
down vote
Another approach is to use the am-gm inequality As Siong Thye Goh has already mentioned, there is a typo in the exercise.
The i´th summand of the LHS is $sqrt[a_i-(i-1)]cdot 1$
Applying am-gm we get $sqrt[a_i-(i-1)]cdot 1leq fraca_i-i+1+12=fraca_i-i+22$
Now we can sum up the terms:
$$sum_i=1^n fraca_i-i+22=-fracn(n-3)2+sum_i=1^n fraca_i2$$
Now we have the following statement:
The equality holds if $b_1=b_2=ldots =b_nquad forall b_i in mathbb R^+$
Consequently the equality is true if $a_n=n, a_n-1=n-1, ldots, a_1=1, $
answered Jul 24 at 17:49
callculus
16.4k31427
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You are indeed given that
$$sum_i=1^na_i=fracn(n-3)2$$
and upon substituting $n=100$, we get
$$frac100(97)2=4850$$
It is correct.
Remark:
The correction is the second equality should be a minus sign, from there we can prove that $a_n=n$ and hence the sum of the first $100$ positive integers is $5050$.
That is the actual question is
Let $a_1, a_2, ldots , a_n$ be real numbers such that
beginalign&sqrta_1+sqrta_2-1+sqrta_3-2+ldots + sqrta_n-(n-1) =\
½ (a_1+a_2+ldots+a_n)colorred-fracn(n-3)4endalign
Compute the value of $sum_i=1^100a_i$.
answered Jul 24 at 16:30
Siong Thye Goh
77.4k134795
by this method answer must be 4850 but answer in my textbook is 5050
– Rafael Nadal
Jul 24 at 16:33
which book are you using? regardless of which method, all correct methods should lead to the same solution.
– Siong Thye Goh
Jul 24 at 16:34
sir i can send u snapshot of its solution please clarify my doubt
– Rafael Nadal
Jul 24 at 16:36
1
most likely it is an error in the book solution. there are mistakes in books.
– Siong Thye Goh
Jul 24 at 16:37
sir tell me anyplace where i can send u snapshot of solution given in my book
– Rafael Nadal
Jul 24 at 16:38
 |Â
show 2 more comments
up vote
3
down vote
accepted
You are indeed given that
$$sum_i=1^na_i=fracn(n-3)2$$
and upon substituting $n=100$, we get
$$frac100(97)2=4850$$
It is correct.
Remark:
The correction is the second equality should be a minus sign, from there we can prove that $a_n=n$ and hence the sum of the first $100$ positive integers is $5050$.
That is the actual question is
Let $a_1, a_2, ldots , a_n$ be real numbers such that
beginalign&sqrta_1+sqrta_2-1+sqrta_3-2+ldots + sqrta_n-(n-1) =\
½ (a_1+a_2+ldots+a_n)colorred-fracn(n-3)4endalign
Compute the value of $sum_i=1^100a_i$.
answered Jul 24 at 16:30
Siong Thye Goh
77.4k134795
by this method answer must be 4850 but answer in my textbook is 5050
– Rafael Nadal
Jul 24 at 16:33
which book are you using? regardless of which method, all correct methods should lead to the same solution.
– Siong Thye Goh
Jul 24 at 16:34
sir i can send u snapshot of its solution please clarify my doubt
– Rafael Nadal
Jul 24 at 16:36
1
most likely it is an error in the book solution. there are mistakes in books.
– Siong Thye Goh
Jul 24 at 16:37
sir tell me anyplace where i can send u snapshot of solution given in my book
– Rafael Nadal
Jul 24 at 16:38
 |Â
show 2 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You are indeed given that
$$sum_i=1^na_i=fracn(n-3)2$$
and upon substituting $n=100$, we get
$$frac100(97)2=4850$$
It is correct.
Remark:
The correction is the second equality should be a minus sign, from there we can prove that $a_n=n$ and hence the sum of the first $100$ positive integers is $5050$.
That is the actual question is
Let $a_1, a_2, ldots , a_n$ be real numbers such that
beginalign&sqrta_1+sqrta_2-1+sqrta_3-2+ldots + sqrta_n-(n-1) =\
½ (a_1+a_2+ldots+a_n)colorred-fracn(n-3)4endalign
Compute the value of $sum_i=1^100a_i$.
answered Jul 24 at 16:30
Siong Thye Goh
77.4k134795
You are indeed given that
$$sum_i=1^na_i=fracn(n-3)2$$
and upon substituting $n=100$, we get
$$frac100(97)2=4850$$
It is correct.
Remark:
The correction is the second equality should be a minus sign, from there we can prove that $a_n=n$ and hence the sum of the first $100$ positive integers is $5050$.
That is the actual question is
Let $a_1, a_2, ldots , a_n$ be real numbers such that
beginalign&sqrta_1+sqrta_2-1+sqrta_3-2+ldots + sqrta_n-(n-1) =\
½ (a_1+a_2+ldots+a_n)colorred-fracn(n-3)4endalign
Compute the value of $sum_i=1^100a_i$.
answered Jul 24 at 16:30
Siong Thye Goh
77.4k134795
edited Jul 24 at 17:43
answered Jul 24 at 16:30
Siong Thye Goh
77.4k134795
answered Jul 24 at 16:30
Siong Thye Goh
77.4k134795
answered Jul 24 at 16:30
Siong Thye Goh
77.4k134795
77.4k134795
by this method answer must be 4850 but answer in my textbook is 5050
– Rafael Nadal
Jul 24 at 16:33
which book are you using? regardless of which method, all correct methods should lead to the same solution.
– Siong Thye Goh
Jul 24 at 16:34
sir i can send u snapshot of its solution please clarify my doubt
– Rafael Nadal
Jul 24 at 16:36
1
most likely it is an error in the book solution. there are mistakes in books.
– Siong Thye Goh
Jul 24 at 16:37
sir tell me anyplace where i can send u snapshot of solution given in my book
– Rafael Nadal
Jul 24 at 16:38
 |Â
show 2 more comments
by this method answer must be 4850 but answer in my textbook is 5050
– Rafael Nadal
Jul 24 at 16:33
which book are you using? regardless of which method, all correct methods should lead to the same solution.
– Siong Thye Goh
Jul 24 at 16:34
sir i can send u snapshot of its solution please clarify my doubt
– Rafael Nadal
Jul 24 at 16:36
1
most likely it is an error in the book solution. there are mistakes in books.
– Siong Thye Goh
Jul 24 at 16:37
sir tell me anyplace where i can send u snapshot of solution given in my book
– Rafael Nadal
Jul 24 at 16:38
by this method answer must be 4850 but answer in my textbook is 5050
– Rafael Nadal
Jul 24 at 16:33
by this method answer must be 4850 but answer in my textbook is 5050
– Rafael Nadal
Jul 24 at 16:33
which book are you using? regardless of which method, all correct methods should lead to the same solution.
– Siong Thye Goh
Jul 24 at 16:34
which book are you using? regardless of which method, all correct methods should lead to the same solution.
– Siong Thye Goh
Jul 24 at 16:34
sir i can send u snapshot of its solution please clarify my doubt
– Rafael Nadal
Jul 24 at 16:36
sir i can send u snapshot of its solution please clarify my doubt
– Rafael Nadal
Jul 24 at 16:36
1
1
most likely it is an error in the book solution. there are mistakes in books.
– Siong Thye Goh
Jul 24 at 16:37
most likely it is an error in the book solution. there are mistakes in books.
– Siong Thye Goh
Jul 24 at 16:37
sir tell me anyplace where i can send u snapshot of solution given in my book
– Rafael Nadal
Jul 24 at 16:38
sir tell me anyplace where i can send u snapshot of solution given in my book
– Rafael Nadal
Jul 24 at 16:38
 |Â
show 2 more comments
up vote
1
down vote
Another approach is to use the am-gm inequality As Siong Thye Goh has already mentioned, there is a typo in the exercise.
The i´th summand of the LHS is $sqrt[a_i-(i-1)]cdot 1$
Applying am-gm we get $sqrt[a_i-(i-1)]cdot 1leq fraca_i-i+1+12=fraca_i-i+22$
Now we can sum up the terms:
$$sum_i=1^n fraca_i-i+22=-fracn(n-3)2+sum_i=1^n fraca_i2$$
Now we have the following statement:
The equality holds if $b_1=b_2=ldots =b_nquad forall b_i in mathbb R^+$
Consequently the equality is true if $a_n=n, a_n-1=n-1, ldots, a_1=1, $
answered Jul 24 at 17:49
callculus
16.4k31427
add a comment |Â
up vote
1
down vote
Another approach is to use the am-gm inequality As Siong Thye Goh has already mentioned, there is a typo in the exercise.
The i´th summand of the LHS is $sqrt[a_i-(i-1)]cdot 1$
Applying am-gm we get $sqrt[a_i-(i-1)]cdot 1leq fraca_i-i+1+12=fraca_i-i+22$
Now we can sum up the terms:
$$sum_i=1^n fraca_i-i+22=-fracn(n-3)2+sum_i=1^n fraca_i2$$
Now we have the following statement:
The equality holds if $b_1=b_2=ldots =b_nquad forall b_i in mathbb R^+$
Consequently the equality is true if $a_n=n, a_n-1=n-1, ldots, a_1=1, $
answered Jul 24 at 17:49
callculus
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Another approach is to use the am-gm inequality As Siong Thye Goh has already mentioned, there is a typo in the exercise.
The i´th summand of the LHS is $sqrt[a_i-(i-1)]cdot 1$
Applying am-gm we get $sqrt[a_i-(i-1)]cdot 1leq fraca_i-i+1+12=fraca_i-i+22$
Now we can sum up the terms:
$$sum_i=1^n fraca_i-i+22=-fracn(n-3)2+sum_i=1^n fraca_i2$$
Now we have the following statement:
The equality holds if $b_1=b_2=ldots =b_nquad forall b_i in mathbb R^+$
Consequently the equality is true if $a_n=n, a_n-1=n-1, ldots, a_1=1, $
answered Jul 24 at 17:49
callculus
16.4k31427
Another approach is to use the am-gm inequality As Siong Thye Goh has already mentioned, there is a typo in the exercise.
The i´th summand of the LHS is $sqrt[a_i-(i-1)]cdot 1$
Applying am-gm we get $sqrt[a_i-(i-1)]cdot 1leq fraca_i-i+1+12=fraca_i-i+22$
Now we can sum up the terms:
$$sum_i=1^n fraca_i-i+22=-fracn(n-3)2+sum_i=1^n fraca_i2$$
Now we have the following statement:
The equality holds if $b_1=b_2=ldots =b_nquad forall b_i in mathbb R^+$
Consequently the equality is true if $a_n=n, a_n-1=n-1, ldots, a_1=1, $
answered Jul 24 at 17:49
callculus
16.4k31427
answered Jul 24 at 17:49
callculus
16.4k31427
answered Jul 24 at 17:49
callculus
16.4k31427
answered Jul 24 at 17:49
callculus
16.4k31427
16.4k31427
add a comment |Â
add a comment |Â
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1
Looks to me like the textbook is in error. $sum_n=1^100n=5050$ may have something to do with the mistake.
– saulspatz
Jul 24 at 16:35
1
You can type the solution in your question.
– saulspatz
Jul 24 at 16:42
2
No. You should type your question and not post images. If it's not worth your time and trouble to type the question, why is it worth my time and trouble to type an answer?
– saulspatz
Jul 24 at 16:54
2
Read my last comment. Stop sending me messages. I don't want to chat with you.
– saulspatz
Jul 24 at 17:05
2
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have included some text from the pictures into your post. Please, edit it further if needed.
– Martin Sleziak
Jul 25 at 8:42