Math Problem Statement

Travailler moi cet exercice par le programme éducatif d un lyceean

Soit (u_n) une suite définie sur IN*. On pose S_n = u_1 + u_2 + ... + u_n et on suppose que S_n = 3n^2 + 5n. 1) Calculer u_1, u_2 et u_3. 2) a. Exprimer S_{n-1} en fonction de n b. Exprimer u_n en fonction de n 3) Montrer que (u_n) est une suite arithmétique. 4) Représenter (u_n) et (S_n).

Solution

Let's break down and solve the exercise in parts. Here's the step-by-step solution to each of the questions in the problem:


Given Information:

We are given a sequence (un)(u_n) defined on N\mathbb{N}^*, and the sum of the first nn terms is given by:

Sn=u1+u2++unS_n = u_1 + u_2 + \dots + u_n with Sn=3n2+5nS_n = 3n^2 + 5n.

1) Calculate u1u_1, u2u_2, and u3u_3:

To calculate the individual terms u1u_1, u2u_2, and u3u_3, we need to recognize that:

un=SnSn1u_n = S_n - S_{n-1}

Calculation of u1u_1:

Since S1=3(1)2+5(1)=3+5=8S_1 = 3(1)^2 + 5(1) = 3 + 5 = 8, we have u1=S1=8u_1 = S_1 = 8.

Calculation of u2u_2:

We need to compute S2S_2 and S1S_1 first:

S2=3(2)2+5(2)=12+10=22S_2 = 3(2)^2 + 5(2) = 12 + 10 = 22 Thus: u2=S2S1=228=14u_2 = S_2 - S_1 = 22 - 8 = 14

Calculation of u3u_3:

Similarly, we calculate S3S_3:

S3=3(3)2+5(3)=27+15=42S_3 = 3(3)^2 + 5(3) = 27 + 15 = 42 Thus: u3=S3S2=4222=20u_3 = S_3 - S_2 = 42 - 22 = 20

Summary:

  • u1=8u_1 = 8
  • u2=14u_2 = 14
  • u3=20u_3 = 20

2a) Express Sn1S_{n-1} as a function of nn:

Since Sn=3n2+5nS_n = 3n^2 + 5n, we can substitute n1n-1 in place of nn to find Sn1S_{n-1}:

Sn1=3(n1)2+5(n1)S_{n-1} = 3(n-1)^2 + 5(n-1) Expanding this expression:

Sn1=3(n22n+1)+5(n1)S_{n-1} = 3(n^2 - 2n + 1) + 5(n-1) Sn1=3n26n+3+5n5S_{n-1} = 3n^2 - 6n + 3 + 5n - 5 Sn1=3n2n2S_{n-1} = 3n^2 - n - 2


2b) Express unu_n as a function of nn:

Now, using the definition of un=SnSn1u_n = S_n - S_{n-1}:

un=(3n2+5n)(3n2n2)u_n = (3n^2 + 5n) - (3n^2 - n - 2) Simplifying the expression:

un=3n2+5n3n2+n+2u_n = 3n^2 + 5n - 3n^2 + n + 2 un=6n+2u_n = 6n + 2

Thus, the general term of the sequence is:

un=6n+2u_n = 6n + 2


3) Show that (un)(u_n) is an arithmetic sequence:

We have found that un=6n+2u_n = 6n + 2. This is clearly an arithmetic sequence, as it is in the form un=u1+(n1)×ru_n = u_1 + (n-1) \times r, where:

  • The first term u1=8u_1 = 8,
  • The common difference (reason) r=6r = 6 (since un+1un=6u_{n+1} - u_n = 6).

Thus, the sequence (un)(u_n) is an arithmetic sequence with:

  • First term u1=8u_1 = 8,
  • Common difference r=6r = 6.

4) Graphical representation:

The final part asks for a graphical representation of the sequences (un)(u_n) and (Sn)(S_n) on the same coordinate system.

  • The sequence un=6n+2u_n = 6n + 2 is linear, with slope 6.
  • The sequence Sn=3n2+5nS_n = 3n^2 + 5n is quadratic, with a parabolic shape opening upwards.

We can plot both of these sequences for values of nNn \in \mathbb{N}^*, but the detailed plot would typically be done using software or graphing tools.


Questions to Expand Understanding:

  1. What is the general formula for the sum of an arithmetic sequence, and how would you apply it to unu_n?
  2. How does the sequence SnS_n grow compared to unu_n as nn becomes large?
  3. Can you find the value of u4u_4 using both the recursive and the explicit formula?
  4. What is the significance of the common difference rr in an arithmetic sequence?
  5. How would the results change if SnS_n were a cubic function instead of quadratic?

Tip: In problems involving sequences and sums, finding the difference between consecutive terms is a powerful way to identify patterns, especially for arithmetic or geometric sequences.

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Math Problem Analysis

Mathematical Concepts

Arithmetic sequences
Quadratic functions
Summation of sequences

Formulas

S_n = 3n^2 + 5n
u_n = S_n - S_{n-1}
Arithmetic sequence formula: u_n = u_1 + (n-1)r

Theorems

Arithmetic progression
Difference of sums to find terms in a sequence

Suitable Grade Level

Grades 10-12