How do I find the General Term of a Sequence?

XEROX_FAN
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1. Write down the general term of each of the following sequences in simplest form:
So far I have been using guess and check to find the solutions, but it doesn't seem much use when you get to a sequence more complicated like this: 1, -4, 7, -10, 13, ...

I really need to find an equation that I can use to determine the general term for any given sequence.

Homework Equations

?[/B] Umm trying to find one ...

The Attempt at a Solution

:[/B] tn = -n
and that's about as far as I got ...
 
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XEROX_FAN said:
1. Write down the general term of each of the following sequences in simplest form:
So far I have been using guess and check to find the solutions, but it doesn't seem much use when you get to a sequence more complicated like this: 1, -4, 7, -10, 13, ...

I really need to find an equation that I can use to determine the general term for any given sequence.2. Homework Equations ? Umm trying to find one ...
3. The Attempt at a Solution : tn = -n
and that's about as far as I got ...

There really is no general formula to find the general term of a sequence. You essentially have to look for a pattern in the numbers.

Notice the terms in your sequence are alternating positive and negative. This tells you ##(-1)^n## will be part of the general term. What else do you notice?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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