# How do you find the sum of the series #i^2# from i=1 to 12?

We have the formula

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Here's a non-standard way to do it without having to remember individual formulae for different kinds of sequences...

The sequence of squares looks like this:

The sequence of differences of that sequence is the sequence of squares:

The sequence of differences of the sequence of squares is a linear (arithmetic) sequence:

The sequence of differences of that sequence is a constant sequence:

So:

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To find the sum of the series (i^2) from (i=1) to (12), you can use the formula for the sum of the first (n) perfect squares, which is (\frac{n(n+1)(2n+1)}{6}). Substituting (n=12) into this formula, you get:

[\frac{12(12+1)(2*12+1)}{6}]

Simplify the expression:

[\frac{12(13)(25)}{6}]

[\frac{3900}{6}]

[650]

So, the sum of the series (i^2) from (i=1) to (12) is (650).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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