Mixing
Sequences and Sums
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There is a list of numbers $a_1 , a_2 , …, a_2010$ . For $1 leq n leq 2010$, where $n$ is positive
integer, let $a_1+a_2+ ldots +a_n = S_n$ . If $a_1 = 2010$ and $S_n = a_nn^2$ for all n, what is
the value of $a_2010$ ?
I've been trying to manipulate the formula but I cant seem to find a good relationship between $a_1$ and $a_2010$ like
$$
a_2010 = fraca_1 +a_2 ... +a_20102010^2
$$
Then tried to use the definition $S_n = a_nn^2 $ over and over again but I cant find a good formula.
sequences-and-series
edited Jul 24 at 13:36
MathOverview
7,95242962
asked Jul 24 at 13:32
SuperMage1
661210
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up vote
4
down vote
favorite
There is a list of numbers $a_1 , a_2 , …, a_2010$ . For $1 leq n leq 2010$, where $n$ is positive
integer, let $a_1+a_2+ ldots +a_n = S_n$ . If $a_1 = 2010$ and $S_n = a_nn^2$ for all n, what is
the value of $a_2010$ ?
I've been trying to manipulate the formula but I cant seem to find a good relationship between $a_1$ and $a_2010$ like
$$
a_2010 = fraca_1 +a_2 ... +a_20102010^2
$$
Then tried to use the definition $S_n = a_nn^2 $ over and over again but I cant find a good formula.
sequences-and-series
edited Jul 24 at 13:36
MathOverview
7,95242962
asked Jul 24 at 13:32
SuperMage1
661210
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
There is a list of numbers $a_1 , a_2 , …, a_2010$ . For $1 leq n leq 2010$, where $n$ is positive
integer, let $a_1+a_2+ ldots +a_n = S_n$ . If $a_1 = 2010$ and $S_n = a_nn^2$ for all n, what is
the value of $a_2010$ ?
I've been trying to manipulate the formula but I cant seem to find a good relationship between $a_1$ and $a_2010$ like
$$
a_2010 = fraca_1 +a_2 ... +a_20102010^2
$$
Then tried to use the definition $S_n = a_nn^2 $ over and over again but I cant find a good formula.
sequences-and-series
edited Jul 24 at 13:36
MathOverview
7,95242962
asked Jul 24 at 13:32
SuperMage1
661210
There is a list of numbers $a_1 , a_2 , …, a_2010$ . For $1 leq n leq 2010$, where $n$ is positive
integer, let $a_1+a_2+ ldots +a_n = S_n$ . If $a_1 = 2010$ and $S_n = a_nn^2$ for all n, what is
the value of $a_2010$ ?
I've been trying to manipulate the formula but I cant seem to find a good relationship between $a_1$ and $a_2010$ like
$$
a_2010 = fraca_1 +a_2 ... +a_20102010^2
$$
Then tried to use the definition $S_n = a_nn^2 $ over and over again but I cant find a good formula.
sequences-and-series
edited Jul 24 at 13:36
MathOverview
7,95242962
asked Jul 24 at 13:32
SuperMage1
661210
edited Jul 24 at 13:36
MathOverview
7,95242962
edited Jul 24 at 13:36
MathOverview
7,95242962
edited Jul 24 at 13:36
MathOverview
7,95242962
7,95242962
asked Jul 24 at 13:32
SuperMage1
661210
asked Jul 24 at 13:32
SuperMage1
661210
asked Jul 24 at 13:32
SuperMage1
661210
661210
add a comment |Â
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
2
down vote
accepted
Hint:
$$S_n-S_n-1=a_n$$
and
$$n^2a_n-(n-1)^2a_n-1=a_n$$
so that
$$a_n=fracn-1n+1a_n-1.$$
Then
$$a_n=fracn-1n+1a_n-1=fracn-1n+1fracn-2n-0a_n-2=fracn-1n+1fracn-2n-0fracn-3n-1a_n-3=fracn-1n+1fracn-2n-0fracn-3n-1fracn-4n-2a_n-4=cdots$$
More generally, after simplification,
$$a_n=frac2n+1frac1n-0a_1=2left(frac1n-frac1n+1right)a_1$$ and the sum telescopes.
answered Jul 24 at 13:54
Yves Daoust
111k665203
Crossed with Siong Thye Goh.
– Yves Daoust
Jul 24 at 13:56
add a comment |Â
up vote
2
down vote
Start from the fact that (for $n>1$) $a_n=S_n-S_n-1$. This means
$$a_n=a_nn^2-a_n-1(n-1)^2.$$
If you use this to express $a_n$ in terms of $a_n-1$, and substitute in the corresponding expression for $a_n-1$, etc, a pattern should emerge.
answered Jul 24 at 13:42
Especially Lime
19k22252
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up vote
2
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Since $a_n=frac1n^2-1sum_k=1^n-1a_k$, the strong induction hypothesis $a_n=frac2a_1n(n+1)$ gives $$a_n=frac2a_1(n-1)(n+1)sum_k=1^n-1(frac1k-frac1k+1)=frac2a_1(n-1)(n+1)(1-frac1n)=frac2a_1n(n+1).$$With our inductive proof complete, $a_2010=frac22011$.
answered Jul 24 at 13:45
J.G.
13.2k11424
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up vote
2
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$$S_n = a_n n^2$$
$$a_n+S_n-1=a_nn^2$$
$$S_n-1=a_n(n^2-1)$$
$$a_n-1(n-1)^2=a_n(n^2-1)$$
$$a_n-1(n-1)=a_n(n+1)$$
$$a_n=fracn-1n+1a_n-1=fracn-1n+1fracn-2na_n-2=fracn-1n+1fracn-2nfracn-3n-1a_n-3=frac(2)(1)(n+1)na_1$$
answered Jul 24 at 13:45
Siong Thye Goh
77.4k134795
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up vote
2
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We have
$$S_2=a_2n^2=4a_2=a_1+a_2 implies a_2=frac13 a_1=670$$
$$S_3=a_3n^2=9a_3=a_1+a_2+a_3 implies 8a_3=frac43 a_1 implies a_3=frac16 a_1=335$$
$$S_4=a_4n^2=16a_4=a_1+a_2+a_3+a_4 implies 15a_4=frac32 a_1 implies a_3=frac110 a_1=201$$
then we claim that $$a_n=frac1T(n)a_1=frac2n(n+1)a_1$$ to be proved by induction, that is
$$S_n+1=a_n+1(n+1)^2=S_n+a_n+1=a_nn^2+a_n+1implies a_n+1((n+1)^2-1)=frac2n^2n(n+1)a_1$$
$$implies a_n+1=frac2n^2n(n+1)(n^2+2n)a_1=frac2(n+1)(n+2)a_1$$
answered Jul 24 at 13:41
gimusi
65.2k73583
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up vote
1
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We have $a_ncdot n^2=a_1+a_2+ldots+a_n=S_n$. Then
$$
beginarraycc
beginarrayrlrl
fracquadquada_1cdot 1^2=&2010\
fracquadquada_2cdot 2^2=&a_1+a_2\
fracquadquada_3cdot 3^2=&a_1+a_2+a_3\
fracquadquada_4cdot 4^2=&a_1+a_2+a_3+a_4\
endarray
&
beginarrayrl
implies a_1=&2010\
implies a_2 =& frac12^2-12010\
implies a_3=&frac13^2-12010+frac13^2-1frac12^2-12010\
implies a_4=¼^2-12010+frac14^2-1frac13^2-12010+frac14^2-1frac13^2-1frac12^2-12010\
endarray
endarray
$$
Now is easy to see
$$
a_n=2010left(frac1n^2-1+frac1n^2-1cdotfrac1(n-1)^2-1+ldots+frac1n^2-1cdotfrac1(n-1)^2-1cdotldotscdotfrac12^2-1right)
$$
answered Jul 24 at 13:41
MathOverview
7,95242962
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint:
$$S_n-S_n-1=a_n$$
and
$$n^2a_n-(n-1)^2a_n-1=a_n$$
so that
$$a_n=fracn-1n+1a_n-1.$$
Then
$$a_n=fracn-1n+1a_n-1=fracn-1n+1fracn-2n-0a_n-2=fracn-1n+1fracn-2n-0fracn-3n-1a_n-3=fracn-1n+1fracn-2n-0fracn-3n-1fracn-4n-2a_n-4=cdots$$
More generally, after simplification,
$$a_n=frac2n+1frac1n-0a_1=2left(frac1n-frac1n+1right)a_1$$ and the sum telescopes.
answered Jul 24 at 13:54
Yves Daoust
111k665203
Crossed with Siong Thye Goh.
– Yves Daoust
Jul 24 at 13:56
add a comment |Â
up vote
2
down vote
accepted
Hint:
$$S_n-S_n-1=a_n$$
and
$$n^2a_n-(n-1)^2a_n-1=a_n$$
so that
$$a_n=fracn-1n+1a_n-1.$$
Then
$$a_n=fracn-1n+1a_n-1=fracn-1n+1fracn-2n-0a_n-2=fracn-1n+1fracn-2n-0fracn-3n-1a_n-3=fracn-1n+1fracn-2n-0fracn-3n-1fracn-4n-2a_n-4=cdots$$
More generally, after simplification,
$$a_n=frac2n+1frac1n-0a_1=2left(frac1n-frac1n+1right)a_1$$ and the sum telescopes.
answered Jul 24 at 13:54
Yves Daoust
111k665203
Crossed with Siong Thye Goh.
– Yves Daoust
Jul 24 at 13:56
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint:
$$S_n-S_n-1=a_n$$
and
$$n^2a_n-(n-1)^2a_n-1=a_n$$
so that
$$a_n=fracn-1n+1a_n-1.$$
Then
$$a_n=fracn-1n+1a_n-1=fracn-1n+1fracn-2n-0a_n-2=fracn-1n+1fracn-2n-0fracn-3n-1a_n-3=fracn-1n+1fracn-2n-0fracn-3n-1fracn-4n-2a_n-4=cdots$$
More generally, after simplification,
$$a_n=frac2n+1frac1n-0a_1=2left(frac1n-frac1n+1right)a_1$$ and the sum telescopes.
answered Jul 24 at 13:54
Yves Daoust
111k665203
Hint:
$$S_n-S_n-1=a_n$$
and
$$n^2a_n-(n-1)^2a_n-1=a_n$$
so that
$$a_n=fracn-1n+1a_n-1.$$
Then
$$a_n=fracn-1n+1a_n-1=fracn-1n+1fracn-2n-0a_n-2=fracn-1n+1fracn-2n-0fracn-3n-1a_n-3=fracn-1n+1fracn-2n-0fracn-3n-1fracn-4n-2a_n-4=cdots$$
More generally, after simplification,
$$a_n=frac2n+1frac1n-0a_1=2left(frac1n-frac1n+1right)a_1$$ and the sum telescopes.
answered Jul 24 at 13:54
Yves Daoust
111k665203
answered Jul 24 at 13:54
Yves Daoust
111k665203
answered Jul 24 at 13:54
Yves Daoust
111k665203
answered Jul 24 at 13:54
Yves Daoust
111k665203
111k665203
Crossed with Siong Thye Goh.
– Yves Daoust
Jul 24 at 13:56
add a comment |Â
Crossed with Siong Thye Goh.
– Yves Daoust
Jul 24 at 13:56
Crossed with Siong Thye Goh.
– Yves Daoust
Jul 24 at 13:56
Crossed with Siong Thye Goh.
– Yves Daoust
Jul 24 at 13:56
add a comment |Â
up vote
2
down vote
Start from the fact that (for $n>1$) $a_n=S_n-S_n-1$. This means
$$a_n=a_nn^2-a_n-1(n-1)^2.$$
If you use this to express $a_n$ in terms of $a_n-1$, and substitute in the corresponding expression for $a_n-1$, etc, a pattern should emerge.
answered Jul 24 at 13:42
Especially Lime
19k22252
add a comment |Â
up vote
2
down vote
Start from the fact that (for $n>1$) $a_n=S_n-S_n-1$. This means
$$a_n=a_nn^2-a_n-1(n-1)^2.$$
If you use this to express $a_n$ in terms of $a_n-1$, and substitute in the corresponding expression for $a_n-1$, etc, a pattern should emerge.
answered Jul 24 at 13:42
Especially Lime
19k22252
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Start from the fact that (for $n>1$) $a_n=S_n-S_n-1$. This means
$$a_n=a_nn^2-a_n-1(n-1)^2.$$
If you use this to express $a_n$ in terms of $a_n-1$, and substitute in the corresponding expression for $a_n-1$, etc, a pattern should emerge.
answered Jul 24 at 13:42
Especially Lime
19k22252
Start from the fact that (for $n>1$) $a_n=S_n-S_n-1$. This means
$$a_n=a_nn^2-a_n-1(n-1)^2.$$
If you use this to express $a_n$ in terms of $a_n-1$, and substitute in the corresponding expression for $a_n-1$, etc, a pattern should emerge.
answered Jul 24 at 13:42
Especially Lime
19k22252
answered Jul 24 at 13:42
Especially Lime
19k22252
answered Jul 24 at 13:42
Especially Lime
19k22252
answered Jul 24 at 13:42
Especially Lime
19k22252
19k22252
add a comment |Â
add a comment |Â
up vote
2
down vote
Since $a_n=frac1n^2-1sum_k=1^n-1a_k$, the strong induction hypothesis $a_n=frac2a_1n(n+1)$ gives $$a_n=frac2a_1(n-1)(n+1)sum_k=1^n-1(frac1k-frac1k+1)=frac2a_1(n-1)(n+1)(1-frac1n)=frac2a_1n(n+1).$$With our inductive proof complete, $a_2010=frac22011$.
answered Jul 24 at 13:45
J.G.
13.2k11424
add a comment |Â
up vote
2
down vote
Since $a_n=frac1n^2-1sum_k=1^n-1a_k$, the strong induction hypothesis $a_n=frac2a_1n(n+1)$ gives $$a_n=frac2a_1(n-1)(n+1)sum_k=1^n-1(frac1k-frac1k+1)=frac2a_1(n-1)(n+1)(1-frac1n)=frac2a_1n(n+1).$$With our inductive proof complete, $a_2010=frac22011$.
answered Jul 24 at 13:45
J.G.
13.2k11424
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since $a_n=frac1n^2-1sum_k=1^n-1a_k$, the strong induction hypothesis $a_n=frac2a_1n(n+1)$ gives $$a_n=frac2a_1(n-1)(n+1)sum_k=1^n-1(frac1k-frac1k+1)=frac2a_1(n-1)(n+1)(1-frac1n)=frac2a_1n(n+1).$$With our inductive proof complete, $a_2010=frac22011$.
answered Jul 24 at 13:45
J.G.
13.2k11424
Since $a_n=frac1n^2-1sum_k=1^n-1a_k$, the strong induction hypothesis $a_n=frac2a_1n(n+1)$ gives $$a_n=frac2a_1(n-1)(n+1)sum_k=1^n-1(frac1k-frac1k+1)=frac2a_1(n-1)(n+1)(1-frac1n)=frac2a_1n(n+1).$$With our inductive proof complete, $a_2010=frac22011$.
answered Jul 24 at 13:45
J.G.
13.2k11424
answered Jul 24 at 13:45
J.G.
13.2k11424
answered Jul 24 at 13:45
J.G.
13.2k11424
answered Jul 24 at 13:45
J.G.
13.2k11424
13.2k11424
add a comment |Â
add a comment |Â
up vote
2
down vote
$$S_n = a_n n^2$$
$$a_n+S_n-1=a_nn^2$$
$$S_n-1=a_n(n^2-1)$$
$$a_n-1(n-1)^2=a_n(n^2-1)$$
$$a_n-1(n-1)=a_n(n+1)$$
$$a_n=fracn-1n+1a_n-1=fracn-1n+1fracn-2na_n-2=fracn-1n+1fracn-2nfracn-3n-1a_n-3=frac(2)(1)(n+1)na_1$$
answered Jul 24 at 13:45
Siong Thye Goh
77.4k134795
add a comment |Â
up vote
2
down vote
$$S_n = a_n n^2$$
$$a_n+S_n-1=a_nn^2$$
$$S_n-1=a_n(n^2-1)$$
$$a_n-1(n-1)^2=a_n(n^2-1)$$
$$a_n-1(n-1)=a_n(n+1)$$
$$a_n=fracn-1n+1a_n-1=fracn-1n+1fracn-2na_n-2=fracn-1n+1fracn-2nfracn-3n-1a_n-3=frac(2)(1)(n+1)na_1$$
answered Jul 24 at 13:45
Siong Thye Goh
77.4k134795
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$S_n = a_n n^2$$
$$a_n+S_n-1=a_nn^2$$
$$S_n-1=a_n(n^2-1)$$
$$a_n-1(n-1)^2=a_n(n^2-1)$$
$$a_n-1(n-1)=a_n(n+1)$$
$$a_n=fracn-1n+1a_n-1=fracn-1n+1fracn-2na_n-2=fracn-1n+1fracn-2nfracn-3n-1a_n-3=frac(2)(1)(n+1)na_1$$
answered Jul 24 at 13:45
Siong Thye Goh
77.4k134795
$$S_n = a_n n^2$$
$$a_n+S_n-1=a_nn^2$$
$$S_n-1=a_n(n^2-1)$$
$$a_n-1(n-1)^2=a_n(n^2-1)$$
$$a_n-1(n-1)=a_n(n+1)$$
$$a_n=fracn-1n+1a_n-1=fracn-1n+1fracn-2na_n-2=fracn-1n+1fracn-2nfracn-3n-1a_n-3=frac(2)(1)(n+1)na_1$$
answered Jul 24 at 13:45
Siong Thye Goh
77.4k134795
answered Jul 24 at 13:45
Siong Thye Goh
77.4k134795
answered Jul 24 at 13:45
Siong Thye Goh
77.4k134795
answered Jul 24 at 13:45
Siong Thye Goh
77.4k134795
77.4k134795
add a comment |Â
add a comment |Â
up vote
2
down vote
We have
$$S_2=a_2n^2=4a_2=a_1+a_2 implies a_2=frac13 a_1=670$$
$$S_3=a_3n^2=9a_3=a_1+a_2+a_3 implies 8a_3=frac43 a_1 implies a_3=frac16 a_1=335$$
$$S_4=a_4n^2=16a_4=a_1+a_2+a_3+a_4 implies 15a_4=frac32 a_1 implies a_3=frac110 a_1=201$$
then we claim that $$a_n=frac1T(n)a_1=frac2n(n+1)a_1$$ to be proved by induction, that is
$$S_n+1=a_n+1(n+1)^2=S_n+a_n+1=a_nn^2+a_n+1implies a_n+1((n+1)^2-1)=frac2n^2n(n+1)a_1$$
$$implies a_n+1=frac2n^2n(n+1)(n^2+2n)a_1=frac2(n+1)(n+2)a_1$$
answered Jul 24 at 13:41
gimusi
65.2k73583
add a comment |Â
up vote
2
down vote
We have
$$S_2=a_2n^2=4a_2=a_1+a_2 implies a_2=frac13 a_1=670$$
$$S_3=a_3n^2=9a_3=a_1+a_2+a_3 implies 8a_3=frac43 a_1 implies a_3=frac16 a_1=335$$
$$S_4=a_4n^2=16a_4=a_1+a_2+a_3+a_4 implies 15a_4=frac32 a_1 implies a_3=frac110 a_1=201$$
then we claim that $$a_n=frac1T(n)a_1=frac2n(n+1)a_1$$ to be proved by induction, that is
$$S_n+1=a_n+1(n+1)^2=S_n+a_n+1=a_nn^2+a_n+1implies a_n+1((n+1)^2-1)=frac2n^2n(n+1)a_1$$
$$implies a_n+1=frac2n^2n(n+1)(n^2+2n)a_1=frac2(n+1)(n+2)a_1$$
answered Jul 24 at 13:41
gimusi
65.2k73583
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have
$$S_2=a_2n^2=4a_2=a_1+a_2 implies a_2=frac13 a_1=670$$
$$S_3=a_3n^2=9a_3=a_1+a_2+a_3 implies 8a_3=frac43 a_1 implies a_3=frac16 a_1=335$$
$$S_4=a_4n^2=16a_4=a_1+a_2+a_3+a_4 implies 15a_4=frac32 a_1 implies a_3=frac110 a_1=201$$
then we claim that $$a_n=frac1T(n)a_1=frac2n(n+1)a_1$$ to be proved by induction, that is
$$S_n+1=a_n+1(n+1)^2=S_n+a_n+1=a_nn^2+a_n+1implies a_n+1((n+1)^2-1)=frac2n^2n(n+1)a_1$$
$$implies a_n+1=frac2n^2n(n+1)(n^2+2n)a_1=frac2(n+1)(n+2)a_1$$
answered Jul 24 at 13:41
gimusi
65.2k73583
We have
$$S_2=a_2n^2=4a_2=a_1+a_2 implies a_2=frac13 a_1=670$$
$$S_3=a_3n^2=9a_3=a_1+a_2+a_3 implies 8a_3=frac43 a_1 implies a_3=frac16 a_1=335$$
$$S_4=a_4n^2=16a_4=a_1+a_2+a_3+a_4 implies 15a_4=frac32 a_1 implies a_3=frac110 a_1=201$$
then we claim that $$a_n=frac1T(n)a_1=frac2n(n+1)a_1$$ to be proved by induction, that is
$$S_n+1=a_n+1(n+1)^2=S_n+a_n+1=a_nn^2+a_n+1implies a_n+1((n+1)^2-1)=frac2n^2n(n+1)a_1$$
$$implies a_n+1=frac2n^2n(n+1)(n^2+2n)a_1=frac2(n+1)(n+2)a_1$$
answered Jul 24 at 13:41
gimusi
65.2k73583
edited Jul 24 at 13:49
answered Jul 24 at 13:41
gimusi
65.2k73583
answered Jul 24 at 13:41
gimusi
65.2k73583
answered Jul 24 at 13:41
gimusi
65.2k73583
65.2k73583
add a comment |Â
add a comment |Â
up vote
1
down vote
We have $a_ncdot n^2=a_1+a_2+ldots+a_n=S_n$. Then
$$
beginarraycc
beginarrayrlrl
fracquadquada_1cdot 1^2=&2010\
fracquadquada_2cdot 2^2=&a_1+a_2\
fracquadquada_3cdot 3^2=&a_1+a_2+a_3\
fracquadquada_4cdot 4^2=&a_1+a_2+a_3+a_4\
endarray
&
beginarrayrl
implies a_1=&2010\
implies a_2 =& frac12^2-12010\
implies a_3=&frac13^2-12010+frac13^2-1frac12^2-12010\
implies a_4=¼^2-12010+frac14^2-1frac13^2-12010+frac14^2-1frac13^2-1frac12^2-12010\
endarray
endarray
$$
Now is easy to see
$$
a_n=2010left(frac1n^2-1+frac1n^2-1cdotfrac1(n-1)^2-1+ldots+frac1n^2-1cdotfrac1(n-1)^2-1cdotldotscdotfrac12^2-1right)
$$
answered Jul 24 at 13:41
MathOverview
7,95242962
add a comment |Â
up vote
1
down vote
We have $a_ncdot n^2=a_1+a_2+ldots+a_n=S_n$. Then
$$
beginarraycc
beginarrayrlrl
fracquadquada_1cdot 1^2=&2010\
fracquadquada_2cdot 2^2=&a_1+a_2\
fracquadquada_3cdot 3^2=&a_1+a_2+a_3\
fracquadquada_4cdot 4^2=&a_1+a_2+a_3+a_4\
endarray
&
beginarrayrl
implies a_1=&2010\
implies a_2 =& frac12^2-12010\
implies a_3=&frac13^2-12010+frac13^2-1frac12^2-12010\
implies a_4=¼^2-12010+frac14^2-1frac13^2-12010+frac14^2-1frac13^2-1frac12^2-12010\
endarray
endarray
$$
Now is easy to see
$$
a_n=2010left(frac1n^2-1+frac1n^2-1cdotfrac1(n-1)^2-1+ldots+frac1n^2-1cdotfrac1(n-1)^2-1cdotldotscdotfrac12^2-1right)
$$
answered Jul 24 at 13:41
MathOverview
7,95242962
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We have $a_ncdot n^2=a_1+a_2+ldots+a_n=S_n$. Then
$$
beginarraycc
beginarrayrlrl
fracquadquada_1cdot 1^2=&2010\
fracquadquada_2cdot 2^2=&a_1+a_2\
fracquadquada_3cdot 3^2=&a_1+a_2+a_3\
fracquadquada_4cdot 4^2=&a_1+a_2+a_3+a_4\
endarray
&
beginarrayrl
implies a_1=&2010\
implies a_2 =& frac12^2-12010\
implies a_3=&frac13^2-12010+frac13^2-1frac12^2-12010\
implies a_4=¼^2-12010+frac14^2-1frac13^2-12010+frac14^2-1frac13^2-1frac12^2-12010\
endarray
endarray
$$
Now is easy to see
$$
a_n=2010left(frac1n^2-1+frac1n^2-1cdotfrac1(n-1)^2-1+ldots+frac1n^2-1cdotfrac1(n-1)^2-1cdotldotscdotfrac12^2-1right)
$$
answered Jul 24 at 13:41
MathOverview
7,95242962
We have $a_ncdot n^2=a_1+a_2+ldots+a_n=S_n$. Then
$$
beginarraycc
beginarrayrlrl
fracquadquada_1cdot 1^2=&2010\
fracquadquada_2cdot 2^2=&a_1+a_2\
fracquadquada_3cdot 3^2=&a_1+a_2+a_3\
fracquadquada_4cdot 4^2=&a_1+a_2+a_3+a_4\
endarray
&
beginarrayrl
implies a_1=&2010\
implies a_2 =& frac12^2-12010\
implies a_3=&frac13^2-12010+frac13^2-1frac12^2-12010\
implies a_4=¼^2-12010+frac14^2-1frac13^2-12010+frac14^2-1frac13^2-1frac12^2-12010\
endarray
endarray
$$
Now is easy to see
$$
a_n=2010left(frac1n^2-1+frac1n^2-1cdotfrac1(n-1)^2-1+ldots+frac1n^2-1cdotfrac1(n-1)^2-1cdotldotscdotfrac12^2-1right)
$$
answered Jul 24 at 13:41
MathOverview
7,95242962
edited Jul 24 at 14:05
answered Jul 24 at 13:41
MathOverview
7,95242962
answered Jul 24 at 13:41
MathOverview
7,95242962
answered Jul 24 at 13:41
MathOverview
7,95242962
7,95242962
add a comment |Â
add a comment |Â
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