AMC 8 Daily Practice Round 9

Complete problem set with solutions and individual problem pages

Problem 10 Hard

What is the value of \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots +\frac{99}{100!} (Factorial can be used in the final result)?

  • A.

    \frac{1}{100!}

  • B.

    \frac{1}{99!}

  • C.

    \frac {1}{100}

  • D.

    1-\frac{1}{100!}

  • E.

    1

Answer:D

\therefore \frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+\cdots +\frac{100-1}{100!}

=\left( \frac{2}{2!}+\frac{3}{3!}+\frac{4}{4!}+\cdots +\frac{100}{100!} \right)-\left( \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots +\frac{1}{100!} \right)

=\left( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots +\frac{1}{99!} \right)-\left( \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots +\frac{1}{100!} \right)

=1-\frac{1}{100!};

\therefore The answer is \text{D}.

Related Problems
Problem 8 AMC 8 Daily Practice Round 9
Problem 9 AMC 8 Daily Practice Round 9
Problem 11 AMC 8 Daily Practice Round 9
Problem 12 AMC 8 Daily Practice Round 9
Think Academy Powered by ChatGPT