Calculus Solutions EX#5.4

Question 1. Evaluate.

(a)

$\begin{aligned} \sum_{k=1}^{3} k^{3}=1^{3}+2^{3}+3^{3}=1+8+27=36 \end{aligned}$

(b)

$\begin{aligned} \sum_{j=2}^{6}(3 j-1)= & (3(2)-1)+(3(3)-1)+(3(4)-1)+(3(5)-1) \\ \\ & +(3(6)-1) \\ \\. = & (6-1)+(9-1)+(12-1)+(15-1)+(18-1) \\ \\ = & 5+8+11+14+17=55 \\ \\ \end{aligned}$

(c)

$\begin{aligned} \sum_{i=-4}^{1}(i^2 -i)= ((-4)^{2}+4)+((-3)^{2}-(-3))+((-2)^{2}-(-2))+\\ \\ ((-1){^2}-(-1))+((0)^{2}-0)+((1)^{2}-1) \\ \\ = (16+4)+(9+3)+(4+2)+(1+1)+0+(1-1) \\ \\ = 20+12+6+2=40 \end{aligned}$

(d)

$\begin{aligned} \sum_{n=0}^{5} (1)=1+1+1+1+1+1=6 \end{aligned}$

(e)

$\begin{aligned} \sum_{k=0}^{4}(-2)^{k} & =(-2)^{0}+(-2)^{1}+(-2)^{2}+(-2)^{3}+(-2)^{4} \\ \\ & =1-2+4-8+16=11 \end{aligned}$

(f)

$\begin{aligned} \sum_{n=1}^{6} \sin n \pi=\sin \pi+\sin 2 \pi+\sin 3 \pi+\sin 4 \pi \\+\sin 5 \pi+\sin 6 \pi \\ \\ =0+0+0+0+0+0=0 \end{aligned}$

Question 2.

(a)

$\begin{aligned} \sum_{k=1}^{4} k \sin k\pi/2=\sin \pi/2 +2 \sin 2 \pi/2 +3\sin 3\pi/2 +4\sin 4\pi/2 \\ \\ =\sin \pi/2 +2\sin \pi +3 \sin 3\pi/2 + 4\sin 2\pi \\ \\ = 1+2(0)+ 3(-1)+4(0)=1-3=-2 \end {aligned}$

(b)

$\begin{aligned} \sum_{j=0}^{5} (-1)^j=(-1)^{0}+(-1)^{1}+(-1)^{2}+(-1)^{3}+(-1)^{4}+(-1)^{5} \\ \\ =1-1+1-1+1-1=0 \end{aligned}$

(c)

$\begin{aligned} \sum_{i=7}^{20} {\pi}^2={\pi}^2+{\pi}^2+{\pi}^2+{\pi}^2+{\pi}^2+{\pi}^2+{\pi}^2 \\ \\ +{\pi}^2+{\pi}^2+{\pi}^2+{\pi}^2+{\pi}^2+{\pi}^2+{\pi}^2 \\ \\ =14 {\pi}^2 \end{aligned}$

(d)

$\begin{aligned} \sum_{m=3}^{5} 2^{m+1}=2^{3+1}+ 2^{4+1} +2^{5+1} \\ \\ =2^{4}+2^{5}+2^{6}=16+32+64=112 \end{aligned}$

(e)

$\begin{aligned} \sum_{n=1}^{6} \sqrt{n}=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6} \end{aligned}$

(f)

$\begin{aligned} \sum_{k=0}^{10} \cos k \pi=\cos 0+ \cos \pi + \cos 2 \pi+ \cos 3 \pi + \cos 4 \pi + \cos 5 \pi \\ \\+\cos 6 \pi+\cos 7 \pi+\cos 8 \pi ++ \cos 9 \pi ++ \cos 10 \pi \\ \\ =1-1+1-1+1-1+1-1+1-1+1=1 \end{aligned}$

Questions 3-8. Write each expression in sigma notation but do not evaluate.

$\begin{aligned} 1+2+3+\cdots+10=\sum_{k=1}^{10} k \end{aligned}$

$\begin{aligned} 3 \cdot 1+3 \cdot 2+3 \cdot 3+3 \cdot 4+\cdots+3 \cdot 20 \\ \\ =3 \cdot ( 1+2+3+\cdots+20) \\ \\ =3 \sum_{k=1}^{20} k=\sum_{k=1}^{20} 3 k \end{aligned}$

$\begin{aligned} 2+4+6+8+\ldots+20 \\ \\ =2 \cdot 1+2 \cdot 2+2 \cdot 3+2 \cdot 4+\ldots+2 \cdot 10 \\ \\ =2 \sum_{k=1}^{10} k=\sum_{k=1}^{10} 2 k \end{aligned}$

$\begin{aligned} 1+3+5+7+\cdots+15 \\ \\ \text {We shall find its general term} \\ \\ & a=1, \quad d=3-1=2 \\ \\ & a_{k}=a+(k-1) d \\ \\ & a_{k}=1+(k-1) 2=1+2 k-2=2 k-1 \\ \\ & 1+3+5+7+\cdots+15=\sum_{k=1}^{8}(2 k-1) \end{aligned}$

$\begin{aligned} 1-3+5-7+9-11=(-1)^{2} 1+(-1)^{3} 3+(-1)^{4} 5+(-1)^{5} 7+(-1)^{6} 9+(-1)^{7} 11 \\ \\ \text { This is an odd series with alternating (+) and (-)} \\ \\ \therefore \quad 1-3+5-7+9-11=\sum_{k=1}^{6}(-1)^{k+1}(2 k-1) \end{aligned}$

$\begin{aligned} 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5} \\ \\ =(-1)^{2} 1+(-1)^{3} \frac{1}{2}+(-1)^{4} \frac{1}{3}+(-1)^{5} \frac{1}{4}+(-1)^{6} \frac{1}{5} \\ \\ =\sum_{k=1}^{5}(-1)^{k+1} \frac{1}{k} \end{aligned}$

9(a) Express the sum of the even integers from 2 to 100 in sigma notation.

$2+4+6+8+\cdots+100$

$\begin{aligned} a=2, \quad d & =4-2=2 \\ \\ \text { General term, } a_{k} & =a+(k-1) d \\ \\ a_{k} & =2+(2+) \\ \\ a_{k} & =2+(k-1) 2=2+2 k-2=2 k \\ \\ \therefore \quad & 2+4+6+8+\cdots+100= \sum_{k=1}^{50} 2 k \end{aligned}$

9(b) Express the sum of the odd integers from 1 to 99 in Sigma notation.

$\begin{aligned} 1+3+5+7+\cdots+99 \\ \\ a=1, \quad d=3-1=2 \\ \\ \text {General term, } a_{k}=a+(k-1) d \\ \\ a_{k}=1+(k-1) 2=1+2 k-2=2 k-1 \\ \\ \therefore \quad 1+3+5+7+\cdots+99=\sum_{k=1}^{50}(2 k-1) \end{aligned}$

10(a). Write in sigma notation

$\begin{aligned} & a_{1}-a_{2}+a_{3}-a_{4}+a_{5} \\ \\ & =(-1)^{2} a_{1}+(-1)^{3} a_{2}+(-1)^{4} a_{3}+(-1)^{5}a_{4}+(-1)^{6} a_{5} \\ \\ & =\sum_{k=1}^{5}(-1)^{k+1} a_{k} \end{aligned}$

10(b)

$\begin{aligned} & -b_{0}+b_{1}-b_{2}+b_{3}-b_{4}+b_{5} \\ \\ & \text {This can be written as } \\ \\ & (-1)^{1} b_{0}+(-1)^{2} b_{1}+(-1)^{3} b_{2}+(-1)^{4}b_{3} \\ \\ & +(-1)^{5} b_{4}+(-1)^{6} b_{5} \\ \\ & =\sum_{k=0}^{5}(-1)^{k+1} b_{k} \end{aligned}$

10(c)

$\begin{aligned} & a_{0}+a_{1} x+a_{2} x^{2}+a_{3 }x^{3}+\cdots+a_{n} x^{n} \\ \\ & \text {This can be written in sigma notation as } \\ \\ & \sum_{k=0}^{n} a_{k} x^{k} \end{aligned}$

10(d)

$\begin{aligned} & a^{5}+a^{4}b+a^{3}b^{2}+a^{2}b^{3}+ab^{4} +b^{5} \end{aligned}$

As we go from left to right , the exponent of $a$ decreases from 5 to 0 and the exponent of $b$ increases from 0 to 5. Therefore in sigma notation above expression can be written as

$\begin{aligned} & =\sum_{k=0}^{5}a^{5-k} b^{k} \end{aligned}$