# Summation and series

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Discrete math questions. Please provide formulas and all calculations for all 22. They are very short answer type questions.

2. Write the first four terms of the sequence defined by bj = 1 + 2j, for all integers j 0.

b0 =

b1 =

b2 =

b3 =

14. Find an explicit formula for the sequence with the given initial terms: 1/4, 2/9, 3/16, 4/25, 5/36, 6/49

23. Compute the summation.

=

27. Write the summation in expanded form.

=

30. Write using summation notation: (13 - 1) + (23 - 1) + (33 - 1) + (43 - 1)

49. 5!/7! =

(Be careful to use parentheses where necessary.)

53. n!/(n 2)! = __________________________________________________

(Be careful to use parentheses where necessary.)

NOTES: I have italicized the variable n (since the Equation Editor normally does this automatically), but it is not necessary. Plain n is fine.

6. (Required) Without using Theorem 4.2.2, use mathematical induction to prove that

P(n): 2 + 4 + 6 + ...+ 2n = n2 + n for all integers n 1.

7. (Required) Without using Theorem 4.2.2, use mathematical induction to prove that

P(n): 1 + 5 + 9 + ... + (4n 3) = n(2n 1) for all integers n 1

Prove the statements in #9-12 by mathematical induction.

9. (Required) P(n): for all integers n 1.

10. (Required) P(n): 13 + 23 + ... + n3 = [n(n + 1)/2]2

11. (Required) P(n): for all integers n 1.

12. (Required) P(n): for each integer n with n 2.

20. (Required) Use the formula for the sum of the first n integers and/or the formula for the sum of a geometric sequence to find the sum 5 + 10 + 15 + 20 + ... + 300.

21. Find an explicit formula for the sequence with the initial terms

2, 6, 12, 20, 30, 42, 56

22. Write using summation notation:

1/2! + 2/3! + 3/4! + ... + n/(n+1)!

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##### Solution Summary

The summation and series in discrete mathematics are examined. There are 22 problems here that deal with summation and series.

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Please see the attachment.

2. Since , then , ,

, .

14. for all integers

23.

27.

30.

49.

53.

6. Proof: is true since . Suppose is true, then we have

. Now we check . By induction, we have

...

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