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January 31st, 2015, 09:33 AM   #1
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Math Focus: Surds!
Nth term! How?

The sequence is:
1, 3, 6, 10, 15, 21, 28

So the interval is:
2, 3, 4, 5, 6, 7

And that interval is:
1

Soo, the nth term may be n^2/2 + n/2

But that seems waaaay to complicated, is there an easier way to, say, work out the 100th term (which was the question)
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January 31st, 2015, 09:42 AM   #2
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This sequence of numbers is famous , they are called triangular numbers.



You don't have to square anything , one of the numbers , n or (n + 1) will be even so divide it by 2 and multiply the result with the other. Just plug and play.

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January 31st, 2015, 09:47 AM   #3
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More explicitly $${n^2 \over 2}+{n \over 2} = \tfrac12(n^2 +n) = \tfrac12n(n+1)$$
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January 31st, 2015, 11:26 AM   #4
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Quote:
Originally Posted by alifeee View Post
The sequence is:
1, 3, 6, 10, 15, 21, 28

So the interval is:
2, 3, 4, 5, 6, 7

And that interval is:
1

Soo, the nth term may be n^2/2 + n/2

But that seems waaaay to complicated, is there an easier way to, say, work out the 100th term (which was the question)
$\displaystyle
\begin{aligned}a_1=1\\
a_2=1+2=3
\\
a_3=1+2+3=6\\
a_4=1+2+3+4=10
\\
.\\.\\.
\\
a_n=1+2+3+ ... +n =\dfrac{n(n+1)}{2}
\end{aligned}
\\\;\\ \\\;\\
a_{100}=\dfrac{100\cdot101}{2}=5050$
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January 31st, 2015, 11:49 AM   #5
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Or, search the oeis; 1,3,6,10,15,21,28 - OEIS
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February 1st, 2015, 09:40 AM   #6
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Quote:
Originally Posted by alifeee View Post
The sequence is:
1, 3, 6, 10, 15, 21, 28


But that seems waaaay to complicated, is there an easier way to, say, work out the 100th term (which was the question)
Yes.

1 = 1 . . . first term

1 + 2 = 3 . . . second term

1 + 2 + 3 = 6 . . . third term

1 + 2 + 3 + 4 = 10 . . . fourth term

.
.
.

Sum = 1 + 2 + 3 + . . . + 98 + 99 + 100 . . . hundredth term
Sum = 100 +99 + 98 + ... + 3 + 2 + 1 . . . . . hundredth term

Twice the sum = 101 + 101 + 101 + ... + 101 + 101 + 101

There are 100 columns.

2*Sum = 100*101

Sum = (100*101)/2


You can finish it.
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