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 October 3rd, 2012, 03:36 AM #1 Newbie   Joined: Oct 2012 Posts: 9 Thanks: 0 Sequence and Series. Summation of series using natural number series... How many terms of the series 1^3 + 2^3 + 3^3 + 4^3 + ... must be taken for the sum to exceed 18 000? Please help.
 October 3rd, 2012, 04:00 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Sequence and Series. $\sum_{k=1}^nk^3>18000=$ $$$\frac{n(n+1)}{2}$$^2>18000$ $\frac{n(n+1)}{2}>60\sqrt{5}$ $n^2+n>120\sqrt{5}>268$ Hence, we find: $16\le n$.
 October 3rd, 2012, 04:29 AM #3 Newbie   Joined: Oct 2012 Posts: 9 Thanks: 0 Re: Sequence and Series. Thanks Mark for answering! But I am confused about how you got your n from the part where n^2 + n > 268 how did you get your n from there? Did you factorise it? if yes, please show me the steps. Thanks.
 October 3rd, 2012, 09:21 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Sequence and Series. No, a bit of mental calculation: $15^2+15=240$ $16^2+16=272$
 October 4th, 2012, 01:35 AM #5 Newbie   Joined: Oct 2012 Posts: 9 Thanks: 0 Re: Sequence and Series. Thanks Mark! =)

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