Answer:
1) True 2) False
Step-by-step explanation:
1) Given [tex] \sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}[/tex]
To verify that the above equality is true or false:
Now find [tex] \sum\limits_{k=0}^8\frac{1}{k+3}[/tex]
Expanding the summation we get
[tex] \sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}[/tex] [tex] \sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}[/tex]
Now find [tex]\sum\limits_{i=3}^{11}\frac{1}{i}[/tex]
Expanding the summation we get
[tex]\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}[/tex]
Comparing the two series we get,
[tex] \sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}[/tex] so the given equality is true.
2) Given [tex] \sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}[/tex]
Verify the above equality is true or false
Now find [tex] \sum\limits_{k=0}^4\frac{3k+3}{k+6}[/tex]
Expanding the summation we get
[tex] \sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}[/tex]
[tex] \sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}[/tex]
now find [tex]\sum\limits_{i=1}^3\frac{3i}{i+5}[/tex]
Expanding the summation we get
[tex]\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}[/tex]
[tex]\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}[/tex]
Comparing the series we get that the given equality is false.
ie, [tex] \sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}[/tex]
