- #1

- 71

- 1

the series: 3 7 12 18 25.

i tried using x(n) = 3 + 4n.

But this doesn't work.. Please help.

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- #1

- 71

- 1

the series: 3 7 12 18 25.

i tried using x(n) = 3 + 4n.

But this doesn't work.. Please help.

- #2

- 24,253

- 16,006

the series: 3 7 12 18 25.

i tried using x(n) = 3 + 4n.

But this doesn't work.. Please help.

I would add ##0## at the beginning and see whether that helps.

- #3

Mentor

- 36,399

- 13,473

Calculate how much the numbers increase in each step. That should lead to some pattern.

- #4

Gold Member

- 2,275

- 2,777

Did you try graphing it?

the series: 3 7 12 18 25.

i tried using x(n) = 3 + 4n.

But this doesn't work.. Please help.

When I graphed your series of numbers, I got a curve.

That implied to me, that "4n", being a linear function, was not correct.

- #5

Gold Member

- 572

- 179

You must find ##f(n)## such that ##a_{n+1}=f(n)+a_{n}## with ##a_{0}=3##, hint: think simple!

Ssnow

Ssnow

- #6

Science Advisor

- 2,277

- 788

Code:

```
3
7 4
12 5 1
18 6 1
25 7 1
```

- #7

- 1

- 0

Regards,

David

- #8

- 68

- 4

This is a recursive formula, that also can be written out as:

x(n) = (2+n) + (2+n-1) + (2+n-2) + (2+n-3) + ... + x(1)

Recognizing the pattern as:

[itex]x(n) = x(1) + \sum_{i=2}^n(2+i) = x(1) - 3 + \sum_{i=1}^n(2+i) = \sum_{i=1}^n2+ \sum_{i=1}^ni = 2n+ n(n+1)/2 = n(n+5)/2 [/itex]

So we find x(n) = n(n+5)/2.

[itex]\sum_{i=1}^nn = n(n+1)/2[/itex] can be found from a table or WolframAlpha(or could also derivated without too much trouble) :)

- #9

- 1

- 0

Type out math instead of posting images.

- #10

##u_{n} = an^{2} + bn + c##

First term: ##u_{1} = a + b + c##

First difference: ##u_{n} - u_{n-1} = 2an + (b-a)##

Second difference: ##(u_{n} - u_{n-1}) - (u_{n-1} - u_{n-2}) = 2a##

So in the case of the sequence ##3,7,12,18,25...## we can write down three relations

##a + b + c = 3##

##3a + b = 4##

##2a = 1##

Which gives ##a = \frac{1}{2}, b = \frac{5}{2}, c= 0##, so

##u_{n} = \frac{1}{2}n^{2} + \frac{5}{2}n##

which is the answer you obtained.

- #11

- 336

- 158

\sum_{n=1}^{a} n^{3} &=& \sum_{n=1}^{a} \bigl( n - 3 n (n + 1) + n (n + 1) (n + 2) \bigr) \\

&=& \frac{1}{2} a (a + 1) - 3 \frac{1}{3} a (a + 1) (a + 2) + \frac{1}{4} a (a + 1) (a + 2) (a + 3) \\

&=&\frac{1}{4} a^{2} (a + 1)^{2} \,.

\end{eqnarray*}$$

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