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The sum of the squares of the first 15 positive integers, [#permalink]
07 Jul 2003, 05:20

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The sum of the squares of the first 15 positive integers, i.e., 1^2 + 2^2 + 3^2 + . . . + 15^2, is equal to 1240. What is the sum of the squares of the second 15 positive integers, i.e., 16^2 + 17^2 + 18^2 + . . . + 30^2?

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

The way I solved a similar problem was to express 16^2 as (15+1)^2 then expand it as (15^2 + 2(15)(1) + 1^2 and similarly 17^2 as (15 + 2)^2 = 15^2 + 2(15)(2) + 2^2, etc.

If you stack up the expansion in the RHS of the above equations, you will note after only two or three of them that you have the addition of 15 constants of 225 in the first term of the expansion, an arithmetic progression from 30 to 450 in the second term, and the sum of squares from 1 to 15 (which is 1240) in the 3rd term. Hence the solution is: 15 x 225 + 15 x (30 + 450) / 2 + 1240 = 8215.

OF course, if you knew the formula for sum of squares from 1 to n, this would be easy (just subtract sum 1 to 15 from 1 to 30), but that is not a formula that the GMAT assumes you know.

BTW, this is admittedly a difficult problem, but my VERY best students have completed this is about 2 minutes. _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

I have a different approach, and a different solution.

Artihmetic mean of 16^2 and 30^2 = (256+900)/2 = 578 number of terms = 15

578*15 = 8670

What is wrong? It took me only 1 minute

Your assumption that 15^2 , 16^2 , 17^2 are in A.P is WRONG ..
Any sequence is said to be in A.P if subsequent terms have a constant /common difference..
So you can not use above approach.. _________________

Hi MBA04,
Well i dont mean to hurt anyone's feelings. I said humor, since the problem didnt seem so striaghtforward also as to say that 15^2, 16^2...etc are in A.P.

I hope you'r aware of progressions. And the above ones are not in A.P. by anyway.

The problem can be solved in a minute, if you know this formula for sum of squares.

I was not hurt, everything is OK.
Thank you very much for your explanation, I understand it perferctly now.
Just one further question, when is AP formula used? (what does AP mean)

I was not hurt, everything is OK. Thank you very much for your explanation, I understand it perferctly now. Just one further question, when is AP formula used? (what does AP mean)

AP is arithmetic Progression - there the elements follow some sort of a rule - such as each is 2 more than the other or such. In this case we can find any member knowing the formula for AP. and we can also find a bunch of other stuff konwing tihs formula; we can figure out the formula by knowing two members and their position numbers (second and third, for example)

In the example with squares, it does not work because the members are not evenly spaced.