What is the sum of the squares of the first n positive integers if the

Say the sum of the squares of the first n positive integers (a.k.a the solution) is x,

1^2 + 2^2 + 3^2 + ....... + n^2 = x

Now, the sum of first n positive integers is A,

2^2 + 4^2 + 6^2 + ....... + (2*n)^2 = A
Take 2^2 (=4) out
4*(1^2 + 2^2 +3^2+ ....... + n^2) = A
Replacing the definition of x from above,
4*x = A
Therefore, x = A/4

Therefore, D. A(1/4)

Let, n = 2

if the sum of the squares of the first n even positive integers is A i.e. \(A = 2^2+4^2 = 20\)

SUm of square of n positive integers \(= 1^2+2^2 = 5\)

\(5 = (1/4)*20\)

i.e. New Sum \(= (1/4)*A\)

Answer: Option D

Solution:

If we let n = 3, we see that the sum of the squares of the first 3 positive integers is 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 and the sum of the squares of the first 3 even positive integers is 2^2 + 4^2 + 6^2 = 4 + 16 + 36 = 56.

We see that the sum of the squares of the first 3 positive integers is ¼ the sum of the squares of the first 3 even positive integers. It should be true that the sum of the squares of the first n positive integers is also ¼ the sum of the squares of the first n even positive integers. Since the latter sum is given as A, therefore the former sum is A(¼).

Alternate Solution:

The nth positive even integer is 2n. We are given that:

2^2 + 4^2 + 6^2 + … + (2n)^2 = A

Let’s factor out 2 from each base:

(2*1)^2 + (2*2)^2 + (2*3)^2 + … + (2 * n)^2 = A

(2^2)*(1^2) + (2^2)*(2^2) + (2^2)*(3^2) + … + (2^2)*(n^2) = A

4(1^2) + 4(2^2) + 4(3^2) + … + 4*(n^2) = A

4(1^2 + 2^2 + 3^2 + … + n^2) = A

1^2 + 2^2 + 3^2 + … + n^2 = A/4

Answer: D

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