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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M7MBA: Moderator; Posts: 2058; Joined: Sun Oct 29, 2017 4:24 amFollowed by: 5 members; Thanked: 1 times

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

Source: Beat The GMAT — Problem Solving | Thu Jul 02, 2020 1:06 am

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

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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