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The sum of the squares of the first \(15\) positive integers \((1^2+2^2+3^2+\cdots+15^2)\) is equal to \(1240.\) What is

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The sum of the squares of the first \(15\) positive integers \((1^2+2^2+3^2+\cdots+15^2)\) is equal to \(1240.\) What is

by M7MBA » Thu Jul 02, 2020 1:06 am

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The sum of the squares of the first \(15\) positive integers \((1^2+2^2+3^2+\cdots+15^2)\) is equal to \(1240.\) What is the sum of the squares of the second \(15\) positive integers \((16^2+17^2+18^2+\cdots+30^2)?\)

(A) \(2480\)
(B) \(3490\)
(C) \(6785\)
(D) \(8215\)
(E) \(9255\)

[spoiler]OA=D[/spoiler]

Source: Manhattan GMAT
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Re: The sum of the squares of the first \(15\) positive integers \((1^2+2^2+3^2+\cdots+15^2)\) is equal to \(1240.\) Wha

by Jay@ManhattanReview » Thu Jul 02, 2020 3:59 am
M7MBA wrote: ↑
Thu Jul 02, 2020 1:06 am
 The sum of the squares of the first \(15\) positive integers \((1^2+2^2+3^2+\cdots+15^2)\) is equal to \(1240.\) What is the sum of the squares of the second \(15\) positive integers \((16^2+17^2+18^2+\cdots+30^2)?\)

(A) \(2480\)
(B) \(3490\)
(C) \(6785\)
(D) \(8215\)
(E) \(9255\)

[spoiler]OA=D[/spoiler]

Source: Manhattan GMAT
We can crack this problem using the formulae.

Sum of the square of first n numbers = [n(n+1)(2n+1)]/6;

So, the sum of the square of first 30 numbers = [30(30+1) (2*30+1)]/6 = 9,455;

=> the sum of the squares of the second \(15\) positive integers \((16^2+17^2+18^2+\cdots+30^2) \)

= [Sum of the square of first 30 numbers – Sum of the square of first 15 numbers]

= 9,455 – 1,240 = 8,215

Correct answer: D

Hope this helps!

-Jay
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