Let (`u_(n)`) be a geometric sequence of common ratio -2, and of first term `u_(0)= -1 `. Let S be the sum of `u_(2)` to `u_(9)`. S=`u_(2)`+`u_(3)`+`u_(4)`+`. . .`+`u_(9)`.

Calculate `u_(2)`

Calculate `u_(9)`.

Deduce S.

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A numerical sequence or a numerical progression is any application of ℕ or a part of ℕ to ℝ

Direction of variation of a sequence: strictly increasing sequence, strictly decreasing sequence.

To say that the sequence (`u_(n)`) is strictly increasing means that:
For any natural number n, `u_(n+1)>u_(n)`

To say that the sequence (`u_(n)`) is strictly decreasing means that:
For any natural number n, `u_(n)>u_(n+1)`.

To show that a sequence is increasing or decreasing:

We can calculate the difference `u_(n+1)-u_(n)`, if this difference is positive then the sequence is increasing, otherwise it is decreasing.

We can also, if the sequence is positive and `u_n!=0`, calculate the ratio `u_(n+1)/u_(n)`, if this ratio is greater than 1 the sequence
is increasing, otherwise it is decreasing.

Arithmetic sequences, geometric sequences

Arithmetic sequences (arithmetic progression)

To say that a sequence (`u_(n)`) is arithmetic means that there is a real r such that for any natural number n, `u_(n+1)`=`u_(n)`+r.
The real r is called the common difference of the sequence (`u_(n)`).
If (`u_(n)`) is an arithmetic sequence of first term `u_(0)`, and common difference r. Then for any natural number n, `u_(n)=u_(0)+nr`

Sum of consecutive terms of an arithmetic sequence

If S=a+...+k is the sum of p consecutive terms of an arithmetic sequence then `S = p(a+k)/2`.
We deduce that `1+2+3+...+n=n(n+1)/2`

Geometric sequences (geometric progression)

To say that a sequence (`u_(n)`) is geometric means that there is a real q such that for any natural n, `u_(n+1)`=`qu_(n)`.
The real q is called the common ratio for the sequence (`u_(n)`).
If (`u_(n)`) is a geometric sequence of first term `u_(0)`, and common ratio q. Then for any natural number n, `u_(n)=u_(0)*q^n`

Sum of consecutive terms of a geometric sequence

If S=a+...+k is the sum of p consecutive terms of a geometric sequence of common ratio q (`q != 1`) then `S = (a-k*q)/(1-q)`.
We deduce that `1+q+q^2+...+q^n=(1-q^(n+1))/(1-q)`