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Number and Algebra - Sequences in general – Series

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A sequence is a list of numbers written in a specific order, where u1,u2,u3,…,unu_1, u_2, u_3, \dots, u_n represent the terms of the sequence.

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The general term unu_n (also called the nthn^{th} term) can be defined by an explicit formula in terms of nn, or by a recursive formula in terms of previous terms (e.g., un=f(un−1)u_n = f(u_{n-1})).

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A series is the sum of the terms of a sequence. The sum of the first nn terms is denoted by Sn=u1+u2+⋯+unS_n = u_1 + u_2 + \dots + u_n.

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Sigma notation ∑\sum is used to write series concisely. The expression ∑r=knur\sum_{r=k}^{n} u_r means the sum of terms from the kthk^{th} term to the nthn^{th} term.

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If the formula for the sum SnS_n is known, the nthn^{th} term can be calculated using the relationship un=Sn−Sn−1u_n = S_n - S_{n-1} for n≥2n \geq 2, and u1=S1u_1 = S_1.

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A sequence is said to be convergent if its terms approach a finite limit as n→∞n \to \infty; otherwise, it is divergent.

📐Formulae

Sn=∑i=1nui=u1+u2+⋯+unS_n = \sum_{i=1}^{n} u_i = u_1 + u_2 + \dots + u_n

un=Sn−Sn−1,n>1u_n = S_n - S_{n-1}, \quad n > 1

∑r=1nc=nc(where c is a constant)\sum_{r=1}^{n} c = nc \quad (\text{where } c \text{ is a constant})

∑r=1n(aur+bvr)=a∑r=1nur+b∑r=1nvr\sum_{r=1}^{n} (au_r + bv_r) = a\sum_{r=1}^{n} u_r + b\sum_{r=1}^{n} v_r

💡Examples

Problem 1:

Evaluate the sum given by ∑r=15(2r−3)\sum_{r=1}^{5} (2r - 3).

Solution:

∑r=15(2r−3)=(2(1)−3)+(2(2)−3)+(2(3)−3)+(2(4)−3)+(2(5)−3)=−1+1+3+5+7=15\begin{aligned} \sum_{r=1}^{5} (2r - 3) &= (2(1)-3) + (2(2)-3) + (2(3)-3) + (2(4)-3) + (2(5)-3) \\ &= -1 + 1 + 3 + 5 + 7 \\ &= 15 \end{aligned}

Explanation:

Substitute each integer value of rr from 11 to 55 into the expression (2r−3)(2r - 3) and add the resulting terms together.

Problem 2:

The sum of the first nn terms of a sequence is given by Sn=3n2−nS_n = 3n^2 - n. Find the 10th10^{th} term, u10u_{10}.

Solution:

u10=S10−S9u_{10} = S_{10} - S_9 S10=3(10)2−10=300−10=290S_{10} = 3(10)^2 - 10 = 300 - 10 = 290 S9=3(9)2−9=3(81)−9=243−9=234S_9 = 3(9)^2 - 9 = 3(81) - 9 = 243 - 9 = 234 u10=290−234=56u_{10} = 290 - 234 = 56

Explanation:

To find a specific term unu_n from a sum formula SnS_n, use the property un=Sn−Sn−1u_n = S_n - S_{n-1}. Here, we calculate S10S_{10} and S9S_9 and find their difference.

Problem 3:

A sequence is defined recursively by u1=4u_1 = 4 and un+1=12un+2u_{n+1} = \frac{1}{2}u_n + 2. Find u3u_3.

Solution:

u2=12u1+2=12(4)+2=2+2=4u_2 = \frac{1}{2}u_1 + 2 = \frac{1}{2}(4) + 2 = 2 + 2 = 4 u3=12u2+2=12(4)+2=2+2=4u_3 = \frac{1}{2}u_2 + 2 = \frac{1}{2}(4) + 2 = 2 + 2 = 4

Explanation:

For recursive sequences, calculate each term sequentially by plugging the previous term into the given recurrence relation.

Sequences in general – Series Grade 11 Notes & Examples