Sum of squares in geometric progression
$begingroup$
In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
and $b_2+b_4=5$. Find the sum of the squares of the first five terms.
If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?
algebra-precalculus geometric-progressions
$endgroup$
add a comment |
$begingroup$
In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
and $b_2+b_4=5$. Find the sum of the squares of the first five terms.
If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?
algebra-precalculus geometric-progressions
$endgroup$
add a comment |
$begingroup$
In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
and $b_2+b_4=5$. Find the sum of the squares of the first five terms.
If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?
algebra-precalculus geometric-progressions
$endgroup$
In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
and $b_2+b_4=5$. Find the sum of the squares of the first five terms.
If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?
algebra-precalculus geometric-progressions
algebra-precalculus geometric-progressions
edited Mar 2 '15 at 6:18
nbubis
27.3k552110
27.3k552110
asked Mar 2 '15 at 4:44
user167045
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.
More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.
$endgroup$
$begingroup$
After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
$endgroup$
– user167045
Mar 2 '15 at 5:04
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@Garvil I don't see any leftover terms.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:10
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Yeah, sorry. My bad.
$endgroup$
– user167045
Mar 2 '15 at 5:12
$begingroup$
No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:13
$begingroup$
Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
$endgroup$
– Janko Bracic
Mar 2 '15 at 5:27
|
show 2 more comments
$begingroup$
Let $b_1=a$, and the common ratio be $r$. Given that
$$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
What is the value of
$$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
Realize that:
$$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
simplifies to the required sum. On substituting:
$$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$
$endgroup$
$begingroup$
Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:09
$begingroup$
20 terms cancel out; remaining five are the required five.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:11
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
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votes
$begingroup$
Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.
More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.
$endgroup$
$begingroup$
After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
$endgroup$
– user167045
Mar 2 '15 at 5:04
$begingroup$
@Garvil I don't see any leftover terms.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:10
$begingroup$
Yeah, sorry. My bad.
$endgroup$
– user167045
Mar 2 '15 at 5:12
$begingroup$
No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:13
$begingroup$
Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
$endgroup$
– Janko Bracic
Mar 2 '15 at 5:27
|
show 2 more comments
$begingroup$
Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.
More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.
$endgroup$
$begingroup$
After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
$endgroup$
– user167045
Mar 2 '15 at 5:04
$begingroup$
@Garvil I don't see any leftover terms.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:10
$begingroup$
Yeah, sorry. My bad.
$endgroup$
– user167045
Mar 2 '15 at 5:12
$begingroup$
No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:13
$begingroup$
Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
$endgroup$
– Janko Bracic
Mar 2 '15 at 5:27
|
show 2 more comments
$begingroup$
Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.
More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.
$endgroup$
Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.
More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.
edited Mar 2 '15 at 5:08
answered Mar 2 '15 at 4:56
Janko BracicJanko Bracic
2,783613
2,783613
$begingroup$
After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
$endgroup$
– user167045
Mar 2 '15 at 5:04
$begingroup$
@Garvil I don't see any leftover terms.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:10
$begingroup$
Yeah, sorry. My bad.
$endgroup$
– user167045
Mar 2 '15 at 5:12
$begingroup$
No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:13
$begingroup$
Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
$endgroup$
– Janko Bracic
Mar 2 '15 at 5:27
|
show 2 more comments
$begingroup$
After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
$endgroup$
– user167045
Mar 2 '15 at 5:04
$begingroup$
@Garvil I don't see any leftover terms.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:10
$begingroup$
Yeah, sorry. My bad.
$endgroup$
– user167045
Mar 2 '15 at 5:12
$begingroup$
No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:13
$begingroup$
Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
$endgroup$
– Janko Bracic
Mar 2 '15 at 5:27
$begingroup$
After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
$endgroup$
– user167045
Mar 2 '15 at 5:04
$begingroup$
After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
$endgroup$
– user167045
Mar 2 '15 at 5:04
$begingroup$
@Garvil I don't see any leftover terms.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:10
$begingroup$
@Garvil I don't see any leftover terms.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:10
$begingroup$
Yeah, sorry. My bad.
$endgroup$
– user167045
Mar 2 '15 at 5:12
$begingroup$
Yeah, sorry. My bad.
$endgroup$
– user167045
Mar 2 '15 at 5:12
$begingroup$
No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:13
$begingroup$
No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:13
$begingroup$
Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
$endgroup$
– Janko Bracic
Mar 2 '15 at 5:27
$begingroup$
Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
$endgroup$
– Janko Bracic
Mar 2 '15 at 5:27
|
show 2 more comments
$begingroup$
Let $b_1=a$, and the common ratio be $r$. Given that
$$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
What is the value of
$$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
Realize that:
$$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
simplifies to the required sum. On substituting:
$$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$
$endgroup$
$begingroup$
Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:09
$begingroup$
20 terms cancel out; remaining five are the required five.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:11
add a comment |
$begingroup$
Let $b_1=a$, and the common ratio be $r$. Given that
$$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
What is the value of
$$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
Realize that:
$$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
simplifies to the required sum. On substituting:
$$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$
$endgroup$
$begingroup$
Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:09
$begingroup$
20 terms cancel out; remaining five are the required five.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:11
add a comment |
$begingroup$
Let $b_1=a$, and the common ratio be $r$. Given that
$$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
What is the value of
$$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
Realize that:
$$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
simplifies to the required sum. On substituting:
$$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$
$endgroup$
Let $b_1=a$, and the common ratio be $r$. Given that
$$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
What is the value of
$$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
Realize that:
$$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
simplifies to the required sum. On substituting:
$$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$
edited Dec 30 '18 at 1:41
Rócherz
2,9863821
2,9863821
answered Mar 2 '15 at 5:08
KugelblitzKugelblitz
3,3351449
3,3351449
$begingroup$
Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:09
$begingroup$
20 terms cancel out; remaining five are the required five.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:11
add a comment |
$begingroup$
Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:09
$begingroup$
20 terms cancel out; remaining five are the required five.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:11
$begingroup$
Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:09
$begingroup$
Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:09
$begingroup$
20 terms cancel out; remaining five are the required five.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:11
$begingroup$
20 terms cancel out; remaining five are the required five.
$endgroup$
– Kugelblitz
Mar 2 '15 at 5:11
add a comment |
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