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# What is the sum of the squares of the first n positive

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What is the sum of the squares of the first n positive [#permalink]  30 Aug 2008, 11:09
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What is the sum of the squares of the first n positive integers if the sum of the squares of the first n even positive integers is A?
a) 3A/4
b) A/2
c) A/3
d) A/4
e) A/8
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Re: Math Question [#permalink]  30 Aug 2008, 11:31
HVD1975 wrote:
What is the sum of the squares of the first n positive integers if the sum of the squares of the first n even positive integers is A?
a) 3A/4
b) A/2
c) A/3
d) A/4
e) A/8

D

just substitute in the number

lets n = 4

1^2 + 2^2 + 3^2 + 4^2 = 30

sum of squares of first 4 even numbers are 2^2 + 4^2 + 6^2 + 8^2 = 120 = A

A/4 = 30
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Re: Math Question [#permalink]  30 Aug 2008, 12:06
Simpler than I thought. Thank you.
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Re: Math Question [#permalink]  30 Aug 2008, 12:25
HVD1975 wrote:
What is the sum of the squares of the first n positive integers if the sum of the squares of the first n even positive integers is A?
a) 3A/4
b) A/2
c) A/3
d) A/4
e) A/8

A =2^2+4^2+...+2n^2 = 4 (1^2+2^2 +...n^2) =

Ans: A/4
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Re: Math Question [#permalink]  31 Aug 2008, 16:53
Suresh, can you please elaborate the equation?
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Re: Math Question [#permalink]  31 Aug 2008, 20:54
subarao wrote:
Suresh, can you please elaborate the equation?

=Sum of the squares of first n even integers.
$$= A =2^2+4^2+...+{(2n)}^2$$
$$= (2^2+2^2 *2^2 +... +2^2*n^2)$$
$$= 2^2(1^2+2^2 +...n^2)$$
$$= 4(1^2+2^2 +...n^2)$$
= 4 * Sum of the squares of first n postive integers.
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Re: Math Question   [#permalink] 31 Aug 2008, 20:54
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