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# SUM of series

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Intern
Joined: 16 Oct 2010
Posts: 7

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21 Oct 2010, 02:43
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Question Stats:

67% (02:13) correct 33% (01:06) wrong based on 1 sessions

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There are 8436 steel balls, each with a radius of 1 centimeter, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is
A. 34 B. 38 C. 36 D. 32

help me out with a solution!
Manager
Joined: 15 Apr 2010
Posts: 173

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21 Oct 2010, 03:56
Anyone with a solution for this problem?

The sequence goes like this:
T(n) = T(n-1) + n

I'm stuck at this point.
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Manager
Joined: 20 Jul 2010
Posts: 51

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21 Oct 2010, 05:49
I think i got it ...

This is what given from the problem:
$$\frac{n(n+1)(n+2)}{6}$$ = 8436
==> n(n+1)(n+2) = 8436 * 6
==> n(n+1)(n+2) = 2 * 2 * 3 * 19 * 37 * 6
==> n(n+1)(n+2) = (2 * 3 * 6) * 37 * (19 * 2)
==> n(n+1)(n+2) = 36 * 37 * 38

So, Ans: 36

By the way, What is radius information doing in this problem?

***
Here is the tedious approach i followed to get $$\frac{n(n+1)(n+2)}{6}$$

Given:
1 + 3 + 6 + 10 + ... nth layer count= 8436
==>
(1) + (1 + 2) + (1 + 2 + 3) + (1+2+3+4) + ... nth layer count= 8436

==> Sum of first 1 numbers + Sum of first 2 numbers + Sum of first 3 numbers + Sum of first 4 numbers + ... + Sum of first (n-1) numbers + Sum of first n numbers = 8436

==> 1 + 3 + 6 + 10 + ... $$\frac{(n-1)n}{2}$$ + $$\frac{n(n+1)}{2}$$ = 8436

Let me reverse the above LHS as below for better reading ...
==> $$\frac{n(n+1)}{2}$$ + $$\frac{(n-1)n}{2}$$ + $$\frac{(n-2) (n-1)}{2}$$ + $$\frac{(n-3) (n-2)}{2}$$ + ... = 8436

==> Now take out the common denominator, then
$$\frac{n(n+1) + (n-1)(n) + (n-2)(n-1) + (n-3)(n-2) + ....}{2}{$$ = 8436

==> (($$n^2$$ + n) + ($$n^2$$ - n) + ($$n^2$$ -3n +2) + ($$n^2$$ -5n + 6) + ....) * (1/2) = 8436

Now add 2 successive terms together ...
i.e, ($$n^2$$ + n) + ($$n^2$$ - n) = $$2n^2$$,
($$n^2$$ -3n +2) + ($$n^2$$ -5n + 6) = 2$$n^2$$ -8n +8

==> ( 2$$n^2$$ + 2$$n^2$$ - 8n +8 + ...) * (1/2) = 8436

Now, cancel out 2 from both numerator and denominator

==> $$n^2$$ + ($$n^2$$ - 4n + 4) + ... = 8436
==> $$n^2$$ + $$(n-1)^2$$ + ... = 8436

Bascially this is what's given in the question:
$$n^2$$ + $$(n-1)^2$$ + $$(n-4)^2$$ + $$(n-6)^2$$ + ... = 8436 ---- (*)

To test the above equation, the sum of first 3 layer counts is:
$$3^2$$ + $$(3-2)^2$$ = 9 + 1 = 10, which is equal to the given numbers 1 + 3 + 6

Now, what (*) formulae represents is the sum of squares of even numbers <= n (where n = last layer)

==> (Sum of squares of first n numbers) - (Sum of squares of odd numbers <= n)
Sum of squares of first n numbers is: ($$\frac{n(n+1)(2n+1)}{6}$$)
Sum of squares of first n odd numbers is: (n)(2n-1)(2n+1)/3. To find out sum of first (n/2) odd numbers substitue n = n/2 in this formulae.

==> ($$\frac{n(n+1)(2n+1)}{6}$$) - ((n/2)*(2*(n/2) - 1)*(2 * (n/2) + 1))/3
==> $$\frac{n(n+1)(n+2)}{6}$$

****

Cheers!
Ravi
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Cheers!
Ravi

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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
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21 Oct 2010, 06:20
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Expert's post

The given series is this: 1, 3, 6, 10, 15....
Note, every nth term is the sum of first n positive integers i.e. 3rd term is 1+2+3 = 6 and so on.

Therefore t(n) = n(n+1)/2 = (n^2)/2 + n/2
Sum of n terms of this series will be S(n) = ∑t(n) = (1/2)∑n^2 + (1/2) ∑n (summation of the terms)
This gives us S(n) = (1/2) n(n+1)(2n+1)/6 +(1/2)n(n+1)/2
Take n(n+1) common from the two terms and simplify the rest to get S(n) = n(n+1)(n+2)/6

Now n(n+1)(n+2)/6 = 8436
or n(n+1)(n+2) = 8436 x 6
At this point, use options. If n = 34, product of 34, 35, 36 will end in 0, not 6, so not possible.
If n = 36, product of 36,37 and 38 will end in 6. This is the only possibility. None of the other products will end in 6 so answer must be 36.
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Manager
Joined: 20 Jul 2010
Posts: 51

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21 Oct 2010, 07:01
Short and Sweet way to derive the formula. Thanks for sharing

Cheers!
Ravi
Intern
Joined: 16 Oct 2010
Posts: 7

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21 Oct 2010, 09:14
many thanks both Ravi & Karishma
Re: SUM of series   [#permalink] 21 Oct 2010, 09:14
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