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.

3.

$$

\begin{aligned}

1+2+3+\cdots+10=\sum_{k=1}^{10} k

\end{aligned}
$$

4.

$$
\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}
$$

5.

$$
\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}
$$

 

6.

$$

\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}

$$

7.

$$

\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}

$$

8.

$$

\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}
$$