Porisma

Worked solution

Fit an exponential-linear recurrence and sum its terms

A sequence satisfies un+1=2un+Anu_{n+1}=2u_n+An for n≥1n\ge1, where AA is constant.

(i) Given u1=5,u2=15u_1=5,u_2=15, find AA and u3u_3.

Its nth term has the form un=a2n+bn+cu_n=a2^n+bn+c, where a,b,ca,b,c are constants.

(ii) Find a,b,ca,b,c.

(iii) Find ∑r=1nur\sum_{r=1}^{n}u_r in terms of nn; you need not simplify your answer.

[2008/I/8]

CJC · Sequences and Series · Example 31 (Self-Reading)

(i) Recurrence constant and third term

Answer

A=5,u3=40A=5,u_3=40.

Substitute the first two terms

u2=2u1+A(1)⟹15=2(5)+A⟹A=5u_2=2u_1+A(1)\quad\Longrightarrow\quad15=2(5)+A\quad\Longrightarrow\quad A=5

The recurrence at n=1 determines A.

Apply the next update

u3=2u2+5(2)=2(15)+10=40\begin{aligned}&u_3\\ & =2u_2+5(2)\\ & =2(15)+10\\ & =40\end{aligned}

The Student copy explicitly prints this calculation; the Teacher copy later uses u3=40 without repeating these lines.

How did this part go?

(ii) Constants in the given explicit form

Answer

a=152,b=−5,c=−5a=\frac{15}{2},b=-5,c=-5.

Build the three simultaneous equations

a(21)+b(1)+c=5⇒2a+b+c=5(1),a(22)+b(2)+c=15⇒4a+2b+c=15(2),a(23)+b(3)+c=40⇒8a+3b+c=40(3).\begin{aligned}a(2^1)+b(1)+c=5&\Rightarrow2a+b+c=5\quad\text{(1)},\\a(2^2)+b(2)+c=15&\Rightarrow4a+2b+c=15\quad\text{(2)},\\a(2^3)+b(3)+c=40&\Rightarrow8a+3b+c=40\quad\text{(3)}.\end{aligned}

Retain the source substitutions at indices 1,2 and 3.

Solve for the constants

a=152,b=−5,c=−5a=\frac{15}{2},\qquad b=-5,\qquad c=-5

The source uses the calculator simultaneous-equation solver. Subtracting consecutive equations also gives 2a+b=10 and 4a+b=25, hence the same constants.

How did this part go?

(iii) Sum through the nth term

Answer

15(2n−1)−5n2(n+1)−5n15(2^n-1)-\frac{5n}{2}(n+1)-5n.

Split geometric, arithmetic and constant sums

∑r=1nur=∑r=1n[152(2r)−5r−5]=152∑r=1n2r−5∑r=1nr−5∑r=1n1.\begin{aligned}\sum_{r=1}^{n}u_r&=\sum_{r=1}^{n}\left[\frac{15}{2}(2^r)-5r-5\right]\\&=\frac{15}{2}\sum_{r=1}^{n}2^r-5\sum_{r=1}^{n}r-5\sum_{r=1}^{n}1.\end{aligned}

The given explicit form separates into three familiar sums.

Identify the correct first terms and counts

152(21+22+⋯+2n)−5(1+2+⋯+n)−5(1+1+⋯+1)\frac{15}{2}(2^1+2^2+\cdots+2^n)-5(1+2+\cdots+n)-5(1+1+\cdots+1)

The geometric powers start at 2 and have ratio 2. The natural numbers start at 1 and have difference 1. Each group has n terms; the source labels incorrectly say GP first term 1 and AP difference 2, although its following substitutions are correct.

Apply the three sum formulas

152[2(2n−1)2−1]−5[n2(n+1)]−5n=15(2n−1)−5n2(n+1)−5n.\begin{aligned}\frac{15}{2}\left[\frac{2(2^n-1)}{2-1}\right]-5\left[\frac n2(n+1)\right]-5n&=15(2^n-1)\\&\quad-\frac{5n}{2}(n+1)-5n.\end{aligned}

The question does not require further simplification. Retain the printed final expression.

How did this part go?

Solution notes

The Student copy prints “in terms of u₁” for part (iii), while the Teacher says “in terms of n”; both worked answers depend on n and u₁ is already fixed at 5. This exercise follows the Teacher ask and retains the Student wording in the native source. The two incorrect series labels are corrected without changing the printed calculation or answer.