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Summation and series: DHS 2024 Promo Q8

Dunman High School2024 PromoPromo P2 Q8●●● Challenging9 marks

What this question tests

Summation and series.

Question

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  1. It is given that ∑r=1nr2=16n(n+1)(2n+1)\displaystyle\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1).

    1. Find ∑r=1n(2r+1+3r−r2)\displaystyle\sum_{r=1}^{n}\left(2^{r+1} + 3r - r^2\right) in the form A(2n−1)+f(n)A(2^n - 1) + \mathrm{f}(n), where AA is a constant and f(n)\mathrm{f}(n) is in fully factorised form. [3]

    2. Using your answer in part a(i), find ∑r=4N(2r+3r−(r−1)2)\displaystyle\sum_{r=4}^{N}\left(2^r + 3r - (r-1)^2\right), leaving your answer in the form B(2N)+C[N(N−1)(5−N)]+DN+EB(2^N) + C\bigl[N(N-1)(5-N)\bigr] + DN + E where BB, CC, DD and EE are constants to be determined. [4]

  2. A sequence is such that v1=pv_1 = p, where pp is a constant, and vn+1=3vn−2,for n≥1.v_{n+1} = 3v_n - 2, \qquad \text{for } n \geq 1. Describe how the sequence behaves when

    1. p=5p = 5, [1]

    2. p=1p = 1. [1]

Show full worked solution▾

(a)(i)

∑r=1n(2r+1+3r−r2)=2∑r=1n2r+3∑r=1nr−∑r=1nr2=2[2(2n−1)2−1]+3[n2(n+1)]−16(n)(n+1)(2n+1)=4(2n−1)+n(n+1)6(9−(2n+1))=4(2n−1)+13n(n+1)(4−n)\begin{aligned} \sum_{r=1}^{n}(2^{r+1}+3r-r^2) &=2\sum_{r=1}^{n}2^r+3\sum_{r=1}^{n}r-\sum_{r=1}^{n}r^2\\ &=2\left[\frac{2(2^n-1)}{2-1}\right] +3\left[\frac n2(n+1)\right] -\frac16(n)(n+1)(2n+1)\\ &=4(2^n-1)+\frac{n(n+1)}{6}(9-(2n+1))\\ &=4(2^n-1)+\frac13n(n+1)(4-n) \end{aligned}

a(ii)

Replace rr with r+1r+1 ∑r=4N(2r+3r−(r−1)2)=∑r+1=4r+1=N(2r+1+3(r+1)−(r+1−1)2)=∑r=3N−1(2r+1+3r−r2)+∑r=3N−13=∑r=1N−1(2r+1+3r−r2)−∑r=12(2r+1+3r−r2)+3(N−1−3+1)=4(2N−1−1)+13(N−1)(N−1+1)(4−(N−1))−(16)+(3N−9)=2(2N)+13N(N−1)(5−N)+3N−29\begin{aligned} \sum_{r=4}^{N}(2^r+3r-(r-1)^2) &=\sum_{r+1=4}^{r+1=N}(2^{r+1}+3(r+1)-(r+1-1)^2)\\ &=\sum_{r=3}^{N-1}(2^{r+1}+3r-r^2)+\sum_{r=3}^{N-1}3\\ &=\sum_{r=1}^{N-1}(2^{r+1}+3r-r^2) -\sum_{r=1}^{2}(2^{r+1}+3r-r^2)+3(N-1-3+1)\\ &=4(2^{N-1}-1)+\frac13(N-1)(N-1+1)(4-(N-1))-(16)+(3N-9)\\ &=2(2^N)+\frac13N(N-1)(5-N)+3N-29 \end{aligned} where B=2,C=13,D=3,E=−29B=2,C=\dfrac13,D=3,E=-29

8b (i, ii)

p=5p=5

Method 1(GC) (preferred) Dunman High School 2024 Promo Q8: solution diagram

Method 2 (algebraic) v2=3v1−2=3(5)−2=13v3=3v2−2=3(13)−2=37v4=3v3−2=3(37)−2=109\begin{aligned} v_2&=3v_1-2=3(5)-2=13\\ v_3&=3v_2-2=3(13)-2=37\\ v_4&=3v_3-2=3(37)-2=109 \end{aligned} …\ldots & so on

∴\thereforeThe sequence increases & diverges.

p=1p=1

Method 1(GC) (preferred) Dunman High School 2024 Promo Q8: solution diagram

Method 2 (algebraic) v2=3v1−2=3(1)−2=1v3=3v2−2=3(1)−2=1v4=3v3−2=3(1)−2=1\begin{aligned} v_2&=3v_1-2=3(1)-2=1\\ v_3&=3v_2-2=3(1)-2=1\\ v_4&=3v_3-2=3(1)-2=1 \end{aligned} …\ldots & so on

∴\thereforeIt is a constant sequence which converges to 1.

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